-
-
- 1 -
- 2 -
- 3 -
- 
-
-
-## 二、程序中的栈
-
-- **函数调用栈**：A(B(C(D())))： 即 A 函数中调用 B，B 调用 C，C 调用 D；在 A 执行的过程中会将 A 压入栈，随后 B 执行时 B 也被压入栈，函数 C 和 D 执行时也会被压入栈。所以当前栈的顺序为：A->B->C->D（栈顶）；函数 D 执行完之后，会弹出栈被释放，弹出栈的顺序为 D->C->B->A;
-- **递归**： 为什么没有停止条件的递归会造成栈溢出？比如函数 A 为递归函数，不断地调用自己（因为函数还没有执行完，不会把函数弹出栈），不停地把相同的函数 A 压入栈，最后造成**栈溢出（Queue OverFloat）**。
-
-## 三、练习
-
-题目：有 6 个元素 6，5，4，3，2，1 按顺序进栈，问下列哪一个不是合法的出栈顺序？
-
-- A：5 4 3 6 1 2 （√）
-- B：4 5 3 2 1 6 （√）
-- C：3 4 6 5 2 1 （×）
-- D：2 3 4 1 5 6 （√）
-
-题目所说的按顺序进栈指的不是一次性全部进栈，而是有进有出，进栈顺序为 6 -> 5 -> 4 -> 3 -> 2 -> 1。
-
-解析：
-
-- A 答案：65 进栈，5 出栈，4 进栈出栈，3 进栈出栈，6 出栈，21 进栈，1 出栈，2 出栈（整体入栈顺序符合 654321）。
-- B 答案：654 进栈，4 出栈，5 出栈，3 进栈出栈，2 进栈出栈，1 进栈出栈，6 出栈（整体的入栈顺序符合 654321）。
-- C 答案：6543 进栈，3 出栈，4 出栈，之后应该 5 出栈而不是 6，所以错误。
-- D 答案：65432 进栈，2 出栈，3 出栈，4 出栈，1 进栈出栈，5 出栈，6 出栈。符合入栈顺序。
-
-## 四、实现栈结构
-
-### 栈常见的操作
-
-- `push()` 添加一个新元素到栈顶位置。
-- `pop()` 移除栈顶的元素，同时返回被移除的元素。
-- `peek()` 返回栈顶的元素，不对栈做任何修改（该方法不会移除栈顶的元素，仅仅返回它）。
-- `isEmpty()` 如果栈里没有任何元素就返回 `true`，否则返回 `false`。
-- `size()` 返回栈里的元素个数。这个方法和数组的 `length` 属性类似。
-- `toString()` 将栈结构的内容以字符串的形式返回。
-
-### 代码实现
-
-```js
-function Stack() {
- this.items = [];
-
- // 1.入栈
- Stack.prototype.push = function (element) {
- return this.items.push(element);
- };
-
- // 2.出栈
- Stack.prototype.pop = function () {
- return this.items.pop();
- };
-
- // 3.查看栈顶元素
- Stack.prototype.peek = function () {
- return this.items[this.items.length - 1];
- };
-
- // 4.判断栈是否为空
- Stack.prototype.isEmpty = function () {
- return this.items.length === 0;
- };
-
- // 5.返回栈的长度
- Stack.prototype.size = function () {
- return this.items.length;
- };
-
- // 6.toString方法
- Stack.prototype.toString = function () {
- // 返回格式如： 10 20 30 等
- var resString = '';
- for (var i = 0; i < this.items.length; i++) {
- resString += this.items[i] + ' ';
- }
- return resString;
- };
-}
-```
-
-### 测试代码
-
-```js
-var stack = new Stack();
-stack.push('10');
-stack.push('20');
-stack.push('30');
-console.log(stack.items); // [ '10', '20', '30' ]
-
-stack.pop();
-console.log(stack.items); // [ '10', '20' ]
-
-console.log(stack.peek()); // 20
-console.log(stack.isEmpty()); // false
-console.log(stack.size()); // 2
-console.log(stack.toString()); // 10 20
-```
-
-## 五、应用
-
-### 1.利用栈结构的特点封装实现十进制转换为二进制的方法
-
-```js
-// 十进制转二进制
-function dec2bin(decNumber) {
- var stack = new Stack();
- while (decNumber > 0) {
- stack.push(decNumber % 2);
- decNumber = Math.floor(decNumber / 2);
- }
-
- var binaryString = '';
- while (!stack.isEmpty()) {
- binaryString += stack.pop();
- }
- return binaryString;
-}
-console.log(dec2bin(100)); // 110010
-```
-
-### 2.[有效的括号](/js-logs/stack#20有效的括号)
-
-- 解题思路：
-- 使用栈结构先进后出的特性
-- 先声明一个栈 stack 变量
-- 遍历整个字符串
-- 发现 ( 、 {、 [ 就入栈
-- 发现 )、}、] 就和栈顶做比较，若栈顶存在对应的括号，将与之对应的括号出栈
-- 遍历结束后，若栈空，则说明所有括号都能找到相匹配的
-
-```js
-/**
- * @param {string} s
- * @return {boolean}
- */
-var isValid = function (s) {
- if (s.length % 2 === 1) {
- return false;
- }
- var stack = [];
- var length = s.length;
- var map = new Map();
- map.set('(', ')');
- map.set('{', '}');
- map.set('[', ']');
-
- for (var i = 0; i < length; i++) {
- var c = s[i];
- if (map.has(c)) {
- stack.push(c);
- } else {
- var stackTop = stack[stack.length - 1];
- if (map.get(stackTop) === c) {
- // 栈顶元素出栈
- stack.pop();
- } else {
- // 栈顶元素与当前元素不匹配 “})” "])" return
- return false;
- }
- }
- }
- return stack.length === 0 ? true : false;
-};
-```
diff --git a/docs/data-structure/02_queue.md b/docs/data-structure/02_queue.md
deleted file mode 100644
index 373059e..0000000
--- a/docs/data-structure/02_queue.md
+++ /dev/null
@@ -1,249 +0,0 @@
----
-nav:
- title: 数据结构
- order: 1
-group:
- title: 常见数据结构
- order: 1
-order: 2
----
-
-# 队列(queue)
-
-## 一、介绍
-
-队列是一种特殊的线性表，生活中类似队列结构的场景：
-
-- 排队，比如在电影院，商场，甚至是厕所排队。
-- 优先排队的人，优先处理。 (买票、结账、WC)。
-
-如图所示：
-
-
- 栈
-
- 
-
-
-队列的限制
-
-- `先进先出` (FIFO：First In First Out)
-- 只允许在表的前端（front）进行删除操作
-- 只允许在表的后端（rear）进行插入操作
-
-如图所示：
-
-
- 生活中的队列
-
- 
-
-
-## 二、程序中的队列
-
-- 打印队列：计算机打印多个文件的时候，需要排队打印
-- 线程队列：当开启多线程时，当新开启的线程所需的资源不足时就先放入线程队列，等待 CPU 处理
-
-## 三、队列的实现
-
-队列有两种实现方式，这里就基于数组方式实现了
-
-- 基于数组实现
-- 基于链表实现
-
-### 队列常见的操作
-
-- `enqueue(element)` 向队列尾部添加一个（或多个）新的项
-- `dequeue()` 移除队列的第一（即排在队列最前面的）项，并返回被移除的元素
-- `front()` 返回队列中的第一个元素——最先被添加，也将是最先被移除的元素。队列不做任何变动（不移除元素，只返回元素信息与 Map 类的 peek 方法非常类似）
-- `isEmpty()` 如果队列中不包含任何元素，返回 true，否则返回 false
-- `size()` 返回队列包含的元素个数，与数组的 length 属性类似
-- `toString()` 将队列中的内容，转成字符串形式
-
-### 代码实现
-
-#### 1.普通队列
-
-```js
-function Queue() {
- this.items = [];
-
- // 1.将元素加入队列
- Queue.prototype.enqueue = function (element) {
- return this.items.push(element);
- };
-
- // 2.从队列中删除元素
- Queue.prototype.dequeue = function () {
- return this.items.shift();
- };
-
- // 3.查看前端的元素
- Queue.prototype.front = function () {
- return this.items[0];
- };
-
- // 4.查看队列是否为空
- Queue.prototype.isEmpty = function () {
- return this.items.length === 0;
- };
-
- // 5.查看队列元素个数
- Queue.prototype.size = function () {
- return this.items.length;
- };
-
- // 6.toString方法
- Queue.prototype.toString = function () {
- // 返回格式如： 10 20 30 等
- var resString = '';
- for (var i = 0; i < this.items.length; i++) {
- resString += this.items[i] + ' ';
- }
- return resString;
- };
-}
-```
-
-#### 测试代码
-
-```js
-var q = new Queue();
-q.enqueue('10');
-q.enqueue('20');
-q.enqueue('30');
-console.log(q.items); // [ '10', '20', '30' ]
-
-q.dequeue();
-console.log(q.items); // [ '20', '30' ]
-
-console.log(q.front()); // 20
-console.log(q.isEmpty()); // false
-console.log(q.size()); // 2
-console.log(q.toString()); // 20 30
-```
-
-#### 2.优先级队列
-
-> 优先级队列，在队列的基础上，增加优先级
->
-> 其实就是在元素`入队时`，`根据传入的优先级，找到合适的位置`，并插入
-
-```js
-function PriorityQueue() {
- // 内部类 也可以用对象存放
- function QueueElement(element, priority) {
- this.element = element;
- this.priority = priority;
- }
-
- this.item = [];
-
- // 1.将元素加入队列
- PriorityQueue.prototype.enqueue = function (element, priority) {
- var queueElement = new QueueElement(element, priority);
- // 如果数组为空，将直接push
- if (this.item.length === 0) {
- this.item.push(queueElement);
- } else {
- var added = false;
- for (var i = 0; i < this.item.length; i++) {
- // 若发现优先级比较大
- if (queueElement.priority < this.item[i].priority) {
- // 插入进去
- this.item.splice(i, 0, queueElement);
- added = true;
- break;
- }
- }
-
- // 若没有执行过插入，说明优先级比较小，直接push
- if (!added) {
- this.item.push(queueElement);
- }
- }
- };
-
- // 2.从队列中删除元素
- PriorityQueue.prototype.dequeue = function () {
- return this.items.shift();
- };
-
- // 3.查看前端的元素
- PriorityQueue.prototype.front = function () {
- return this.items[0];
- };
-
- // 4.查看队列是否为空
- PriorityQueue.prototype.isEmpty = function () {
- return this.items.length === 0;
- };
-
- // 5.查看队列元素个数
- PriorityQueue.prototype.size = function () {
- return this.items.length;
- };
-
- // 6.toString方法
- PriorityQueue.prototype.toString = function () {
- // 返回格式如： 10 20 30 等
- var resString = '';
- for (var i = 0; i < this.items.length; i++) {
- resString += this.items[i] + ' ';
- }
- return resString;
- };
-}
-```
-
-## 四、应用
-
-### 1.击鼓传花
-
-```js
-const Queue = require('./queue');
-
-/**
- * 击鼓传花游戏规则：
- * 几个朋友一起玩一个游戏，围成一圈，开始数数，数到某个数字的人自动淘汰
- * 最后剩下的这个人会获得胜利，请问：最后胜利者原来在哪一个位置上
- */
-let names = ['小明', '小红', '小亮'];
-function passGame(nameList, stopNum = 1) {
- let q = new Queue();
- for (const item of nameList) {
- q.enqueue(item);
- }
-
- while (q.size() > 1) {
- for (let i = 0; i < stopNum - 1; i++) {
- q.enqueue(q.dequeue());
- }
- q.dequeue();
- }
- console.log('最后剩下的人是:', q.front()); // 最后剩下的人是: 小红
- return nameList.indexOf(q.front()); //返回对应的位置
-}
-
-console.log(passGame(names, 3)); // 1
-```
-
-### 2.[最近的请求次数](https://leetcode-cn.com/problems/number-of-recent-calls/)
-
-```js
-var RecentCounter = function () {
- this.queue = [];
-};
-
-/**
- * @param {number} t
- * @return {number}
- */
-RecentCounter.prototype.ping = function (t) {
- this.queue.push(t);
- // 将3000ms之前的数据，踢出队列
- while (this.queue[0] < t - 3000) {
- this.queue.shift();
- }
- // 队列中剩下的元素个数就是最近的请求次数
- return this.queue.length;
-};
-```
diff --git a/docs/data-structure/03_linked_list.md b/docs/data-structure/03_linked_list.md
deleted file mode 100644
index a445e2e..0000000
--- a/docs/data-structure/03_linked_list.md
+++ /dev/null
@@ -1,599 +0,0 @@
----
-nav:
- title: 数据结构
- order: 1
-group:
- title: 常见数据结构
- order: 1
-order: 3
----
-
-# 链表(LinkedList)
-
-## 一、介绍
-
-链表和数组一样，可以用于**存储一系列的元素**，但是链表和数组的实现**机制完全不同**。链表又可以分作**单向链表**和**双向链表**。
-
-链表的每个元素由一个存储**元素本身的节点**和一个**指向下一个元素的引用**（有的语言称为指针或连接）组成。类似于火车头，一节车厢载着乘客（数据），通过节点连接另一节车厢。如下图所示
-
-- 链表的火车结构
-
-
- 队列
-
- 
-
-
-- 链表的数据结构
-
-
- 火车结构
-
- 
-
-
- head 属性指向链表的第一个节点。
-
- 链表中的最后一个节点指向 `null`。
-
- 当链表中一个节点也没有的时候，head 直接指向 `null`。
-
-- 给火车加上数据结构后
-
-
- 链表的数据结构
-
- 
-
-
-### 数组的优缺点
-
-- 优点：
- - 一般存储多个元素时，数组（或列表）可能是最常用的数据结构
-- 缺点：
- - 数组的创建需要申请一段连续的内存空间(一整块内存)，并且大小是固定的，**当前数组不能满足容量需求时，需要扩容**。 (一般情况下是申请一个更大的数组，比如 2 倍，然后将原数组中的元素复制过去)
- - 在数组**开头**或中间位置**插入数据的成本很高**，需要进行**大量元素的位移**
-
-
-
-### 链表的优缺点
-
-- 优点：
- - 链表中的元素在内存中**不必是连续的空间**，可以充分利用计算机的内存，实现灵活的**内存动态管理**。
- - 链表不必在创建时就**确定大小**，并且大小可以**无限地延伸**下去。
- - 链表在**插入和删除**数据时，**时间复杂度**可以达到 O(1)，相对数组效率高很多。
-- 缺点：
- - 链表访问任何一个位置的元素时，都需要**从头开始访问**（无法跳过第一个元素访问任何一个元素）
- - 无法通过下标值直接访问元素，需要从头开始一个个访问，直到找到对应的元素
- - 虽然可以轻松地到达**下一个节点**，但是回到**前一个节点**是很难的
- - 总的来说就是**链表查找效率低**
-
-## 二、链表的实现
-
-### 链表中的常见操作
-
-- `append(element)` 向链表尾部添加一个新的项。
-- `insert(position, element)` 向链表的特定位置插入一个新的项。
-- `get(position)` 获取对应位置的元素。
-- `indexOf(element)` 返回元素在链表中的索引。如果链表中没有该元素就返回-1。
-- `update(position, element)` 修改某个位置的元素。
-- `removeAt(position)` 从链表的特定位置移除一项。
-- `remove(element)` 从链表中移除一项。
-- `isEmpty()` 如果链表中不包含任何元素，返回 true，如果链表长度大于 0 则返回 false。
-- `size()` 返回链表包含的元素个数，与数组的 length 属性类似。
-- `toString()` 由于链表项使用了 Node 类，就需要重写继承自 JavaScript 对象默认的 toString 方法，让其只输出元素的值。
-
-### 1.逐步封装单向链表
-
-#### 1.1 先创建单向链表类 LinkedList，并添加基本属性，再实现单向链表的常用方法：
-
-```js
-function LinkedList() {
- function Node(data) {
- this.data = data;
- this.next = null;
- }
-
- this.head = null;
- this.length = 0;
-}
-```
-
-#### 1.2 实现 append() 方法
-
-```js
-// 1.向链表尾部追加数据
-LinkedList.prototype.append = function (data) {
- var newNode = new Node(data);
- // 若链表为空
- if (this.length === 0) {
- // 将head指向第一个节点
- this.head = newNode;
- } else {
- var current = this.head;
- // 当current.next为null 的时候会退出循环
- while (current.next) {
- // 指针后移
- current = current.next;
- }
- // 当指到最后一个node时，给next append newNode
- current.next = newNode;
- }
- // 注意：append后长度要+1
- this.length += 1;
-};
-```
-
-##### append 过程详解
-
-1. 首先让 `currentNode` 指向第一个节点。
-
-
- 给火车加上数据结构后
-
- 
-
-
-2. 通过 `while` 循环使 `currentNode` 指向最后一个节点，最后通过 `currentNode.next = newNode`，让最后一个节点指向新节点 `newNode`。
-
-
-
- 
-
-
-#### 1.3 实现 toString() 方法
-
-```js
-// 2.方便测试，先完成toString方法
-LinkedList.prototype.toString = function () {
- var current = this.head;
- var resultString = '';
- while (current) {
- resultString += current.data + ' ';
- current = current.next;
- }
- return resultString;
-};
-```
-
-##### 测试代码
-
-```js
-var linked = new LinkedList();
-linked.append('10');
-linked.append('20');
-linked.append('30');
-console.log(linked.toString()); // 10 20 30
-```
-
-#### 1.4 实现 insert()方法
-
-```js
-// 3.插入
-LinkedList.prototype.insert = function (position, data) {
- // 1.对position进行边界情况判断
- if (position < 0 || position > this.length) return false;
-
- var newNode = new Node(data);
- // 2.当在第一个位置插入的时候
- if (position === 0) {
- // 将要新节点的next与原链表相连
- newNode.next = this.head;
- // 改变head的指向到新的节点，有点移花接木的意思了哈～
- this.head = newNode;
- } else {
- // 3.这里要找到合适的插入位置
- var index = 0;
- var current = this.head;
- var previous = null;
-
- // 3.1 遍历的找到要插入的节点位置
- while (index++ < position) {
- previous = current; // 保存前一个节点
- current = current.next; // 保持当前节点
- }
-
- // 3.2 同样是移花接木
- newNode.next = current;
- previous.next = newNode;
- }
-
- this.length += 1;
- return true;
-};
-```
-
-##### 测试代码
-
-```js
-linked.insert(0, '00');
-linked.insert(2, '22');
-linked.insert(5, '55');
-console.log(linked.toString()); // 00 10 22 20 30 55
-```
-
-#### 1.5 实现 get()方法
-
-> 根据位置获取 对应位置的数据
-
-```js
-// 4.获取对应位置的数据
-LinkedList.prototype.get = function (position) {
- if (position < 0 || position >= this.length) return false;
- var current = this.head;
- var index = 0;
- while (index++ < position) {
- current = current.next;
- }
- return current.data;
-};
-```
-
-##### 测试代码
-
-```js
-console.log(linked.get(0)); // 00
-console.log(linked.get(2)); // 22
-console.log(linked.get(5)); // 55
-```
-
-#### 1.6 实现 indexOf()方法
-
-```js
-// 5.返回数据当前位置,若找不到传入数据的位置，返回-1
-LinkedList.prototype.indexOf = function (data) {
- var current = this.head;
- var index = 0;
- while (current) {
- if (current.data === data) return index;
- current = current.next;
- index += 1;
- }
- return -1;
-};
-```
-
-##### 测试代码
-
-```js
-console.log(linked.indexOf('00')); // 0
-console.log(linked.indexOf('22')); // 2
-console.log(linked.indexOf('123123')); // -1
-```
-
-#### 1.7 实现 update()方法
-
-```js
-// 6.更新对应位置对数据
-LinkedList.prototype.update = function (position, newData) {
- if (position < 0 || position >= this.length) return false;
- var current = this.head;
- var index = 0;
- while (index++ < position) {
- current = current.next;
- }
- current.data = newData;
- return true;
-};
-```
-
-##### 测试代码
-
-```js
-linked.update(0, '000');
-console.log(linked.get(0)); // 000
-```
-
-#### 1.8 实现 removeAt()方法
-
-```js
-// 7.根据位置删除数据
-LinkedList.prototype.removeAt = function (position) {
- if (position < 0 || position >= this.length) return null;
- var current = this.head;
-
- if (position === 0) {
- this.head = this.head.next;
- } else {
- var previous = null;
- var index = 0;
- while (index++ < position) {
- previous = current;
- current = current.next;
- }
- // 将前一个节点的next执行下一个节点
- previous.next = current.next;
- }
- this.length -= 1;
- return current.data; // 返回删除的数据
-};
-```
-
-##### 测试代码
-
-```js
-console.log(linked.removeAt(0)); // 000
-console.log(linked.removeAt(4)); // 55
-console.log(linked.toString()); // 10 22 20 30
-```
-
-#### 1.9 实现 remove()方法
-
-```js
-// 8.删除数据
-LinkedList.prototype.remove = function (data) {
- var position = this.indexOf(data);
- return this.removeAt(position);
-};
-```
-
-##### 测试代码
-
-```js
-console.log(linked.remove('10')); // 10
-console.log(linked.toString()); // 22 20 30
-```
-
-### 2.封装完的单向链表
-
-> 包括测试代码和 isEmpty()、size()的实现
-
-```js
-function LinkedList() {
- function Node(data) {
- this.data = data;
- this.next = null;
- }
-
- this.head = null;
- this.length = 0;
-
- // 1.向链表尾部追加数据
- LinkedList.prototype.append = function (data) {
- var newNode = new Node(data);
- // 若链表为空
- if (this.length === 0) {
- // 将head指向第一个节点
- this.head = newNode;
- } else {
- var current = this.head;
- // 当current.next为null 的时候会退出循环
- while (current.next) {
- // 指针后移
- current = current.next;
- }
- // 当指到最后一个node时，给next append newNode
- current.next = newNode;
- }
- // 注意：append后长度要+1
- this.length += 1;
- };
-
- // 3.插入
- LinkedList.prototype.insert = function (position, data) {
- // 1.对position进行边界情况判断
- if (position < 0 || position > this.length) return false;
-
- var newNode = new Node(data);
- // 2.当在第一个位置插入的时候
- if (position === 0) {
- // 将要新节点的next与原链表相连
- newNode.next = this.head;
- // 改变head的指向到新的节点，有点移花接木的意思了哈～
- this.head = newNode;
- } else {
- // 3.这里要找到合适的插入位置
- var index = 0;
- var current = this.head;
- var previous = null;
-
- // 3.1 遍历的找到要插入的节点位置
- while (index++ < position) {
- previous = current; // 保存前一个节点
- current = current.next; // 保持当前节点
- }
-
- // 3.2 同样是移花接木
- newNode.next = current;
- previous.next = newNode;
- }
-
- this.length += 1;
- return true;
- };
-
- // 4.获取对应位置的数据
- LinkedList.prototype.get = function (position) {
- if (position < 0 || position >= this.length) return false;
- var current = this.head;
- var index = 0;
- while (index++ < position) {
- current = current.next;
- }
- return current.data;
- };
-
- // 5.返回数据当前位置
- LinkedList.prototype.indexOf = function (data) {
- var current = this.head;
- var index = 0;
- while (current) {
- if (current.data === data) return index;
- current = current.next;
- index += 1;
- }
- return -1;
- };
-
- // 6.更新对应位置对数据
- LinkedList.prototype.update = function (position, newData) {
- if (position < 0 || position >= this.length) return false;
- var current = this.head;
- var index = 0;
- while (index++ < position) {
- current = current.next;
- }
- current.data = newData;
- return true;
- };
-
- // 7.根据位置删除数据
- LinkedList.prototype.removeAt = function (position) {
- if (position < 0 || position >= this.length) return null;
- var current = this.head;
-
- if (position === 0) {
- this.head = this.head.next;
- } else {
- var previous = null;
- var index = 0;
- while (index++ < position) {
- previous = current;
- current = current.next;
- }
- // 将前一个节点的next执行下一个节点
- previous.next = current.next;
- }
- this.length -= 1;
- return current.data; // 返回删除的数据
- };
-
- // 8.删除数据
- LinkedList.prototype.remove = function (data) {
- var position = this.indexOf(data);
- return this.removeAt(position);
- };
-
- LinkedList.prototype.isEmpty = function () {
- return this.length === 0;
- };
-
- LinkedList.prototype.size = function () {
- return this.length;
- };
-
- // 2.方便测试，先完成toString方法
- LinkedList.prototype.toString = function () {
- var current = this.head;
- var resultString = '';
- while (current) {
- resultString += current.data + ' ';
- current = current.next;
- }
- return resultString;
- };
-}
-
-// 测试代码
-var linked = new LinkedList();
-linked.append('10');
-linked.append('20');
-linked.append('30');
-console.log(linked.toString()); // 10 20 30
-
-linked.insert(0, '00');
-linked.insert(2, '22');
-linked.insert(5, '55');
-console.log(linked.toString()); // 00 10 22 20 30 55
-console.log(linked.get(0)); // 00
-console.log(linked.get(2)); // 22
-console.log(linked.get(5)); // 55
-
-console.log(linked.indexOf('00')); // 0
-console.log(linked.indexOf('22')); // 2
-console.log(linked.indexOf('123123')); // -1
-
-linked.update(0, '000');
-console.log(linked.get(0)); // 000
-
-console.log(linked.removeAt(0)); // 000
-console.log(linked.removeAt(4)); // 55
-console.log(linked.toString()); // 10 22 20 30
-
-console.log(linked.remove('10')); // 10
-console.log(linked.toString()); // 22 20 30
-```
-
-## 三、应用
-
-### 1.[删除链表中的节点](https://leetcode-cn.com/problems/delete-node-in-a-linked-list/)
-
-- 思路
-- 若无任何限制，是一个普通链表，可以用正常方法删除（将上一个节点的 next，指向下一个节点即可）
-- 因为题目限制无法访问链表的头节点
-- 所以需要绕一个弯，将当前节点，指向下一个节点，这个时候当前节点跟下一个节点是同一个节点
-- 然后将下一个节点删除即可，也就是将下个节点的指向下下个节点
-
-```js
-var deleteNode = function (node) {
- node.val = node.next.val;
- node.next = node.next.next;
-};
-```
-
-### 2.[删除排序链表中的重复元素](/js-logs/array#83删除排序链表中的重复元素)
-
-- 思路
-- 因为题目给出的链表是排好序的
-- 所以只需要，遍历链表，判断当前节点与下个节点的值是否相等
-- 遇到相等的情况继续删除，当 遇到 不相等的情况再进行赋值往下走
-
-注意点：
-因为第一次判断的时候用到了 p.next.val,若 p.next 为 null，则会报错，所以 while 中需要考虑 p.next 不为空的情况
-
-```js
-var deleteDuplicates = function (head) {
- var p = head;
- while (p && p.next) {
- if (p.val === p.next.val) {
- p.next = p.next.next;
- } else {
- p = p.next;
- }
- }
- return head;
-};
-```
-
-### 3.[反转链表](/js-logs/linked-list#反转链表)
-
-- 思路
-- 链表太长不容易看出如何反转，所以可以看作两个最短链表反转
-- 短链表反转，只需要将 当前节点的 next 指回 给 上一个节点
-- 所以使用 p1 为头节点，p2 为后节点，双指针一直指回，然后让双指针共同前进即可
-- 因为 p1 最后会指向一个 null，p2 会指向 5
-- 所以最后返回 p2 链表
-
-```js
-var reverseList = function (head) {
- var p1 = head;
- var p2 = null;
- while (p1) {
- var temp = p1.next;
- p1.next = p2;
- p2 = p1;
- p1 = temp;
- }
-
- return p2;
-};
-```
-
-### 4. [环形链表](/js-logs/linked-list#141环形链表)
-
-- 思路
-- 声明快慢两个指针进行赛跑
-- 如果有环，快慢指针肯定会重逢
-- 如果没有环，则不会重逢
-
-```js
-var hasCycle = function (head) {
- var p1 = head;
- var p2 = head;
-
- while (p1 && p2 && p2.next) {
- p1 = p1.next;
- p2 = p2.next.next;
-
- if (p1 === p2) {
- return true;
- }
- }
-
- return false;
-};
-```
diff --git a/docs/data-structure/04_doubly_linked_list.md b/docs/data-structure/04_doubly_linked_list.md
deleted file mode 100644
index a3afdf2..0000000
--- a/docs/data-structure/04_doubly_linked_list.md
+++ /dev/null
@@ -1,883 +0,0 @@
----
-nav:
- title: 数据结构
- order: 1
-group:
- title: 常见数据结构
- order: 1
-order: 4
----
-
-# 双向链表(DoublyLinkedList)
-
-## 一、双向链表的介绍
-
-既可以从头遍历到尾，也可以从**尾遍历到头**。链表相连的过程是**双向**的。实现原理是一个节点**既有向前连接的引用**，也有一个**向后连接的引用**
-
-### 单向链表的特性
-
-- 只能从头遍历到尾或者从尾遍历到头（一般从头到尾）
-- 链表相连的过程是单向的，实现原理是上一个节点中有指向下一个节点的引用
-- 单向链表有一个比较明显的缺点：可以轻松到达下一个节点，但回到前一个节点很难，在实际开发中, 经常会遇到需要回到上一个节点的情况。
-- 由于双向链表的节点指向是双向的，所以双向链表可以有效的解决单向链表存在的问题
-
-### 双向链表结构
-
-
-
- 
-
-
-- 双向链表不仅有**head**指针指向第一个节点，而且有**tail**指针指向最后一个节点
-- 每一个节点由三部分组成：**item**储存数据、**prev**指向前一个节点、**next**指向后一个节点
-- 双向链表的第一个节点的 prev 指向**null**
-- 双向链表的最后一个节点的 next 指向**null**
-
-### 双向链表的缺点
-
-- 每次在**插入或删除**某个节点时，都需要处理四个引用，而不是两个，实现起来会困难些
-- 相对于单向链表，所占**内存空间更大**一些
-
-## 二、双向链表的封装
-
-### 常见操作
-
-- `append(element)` 向链表尾部追加一个新元素
-- `insert(position, element)` 向链表的指定位置插入一个新元素
-- `getElement(position)` 获取指定位置的元素
-- `indexOf(element)` 返回元素在链表中的索引。如果链表中没有该元素就返回 -1
-- `update(position, element)` 修改指定位置上的元素
-- `removeAt(position)` 从链表中的删除指定位置的元素
-- `remove(element)` 从链表删除指定的元素
-- `isEmpty()` 如果链表中不包含任何元素，返回 `true`，如果链表长度大于 0 则返回 `false`
-- `size()` 返回链表包含的元素个数，与数组的 `length` 属性类似
-- `toString()` 由于链表项使用了 Node 类，就需要重写继承自 JavaScript 对象默认的 `toString` 方法，让其只输出元素的值
-- `forwardString()` 返回正向遍历节点字符串形式
-- `backwardString()` 返回反向遍历的节点的字符串形式
-
-### 1.逐步封装双向链表
-
-> 双向链表的插入(insert)和删除(removeAt)方法需要考虑的情况较多，是一个难点
-
-#### 1.1 先创建双向链表类 DoublyLinkedList
-
-创建内部类并添加基本属性，再实现单向链表的常用方法
-
-与单向链表相比
-
-- 双向链表内部类中新添加`this.prev`属性，指向该节点的上一个节点
-- 双向链表新添加 `this.tail` 属性，该属性指向末尾的节点
-
-```js
-function DoublyLinkedList() {
- // 内部类
- function Node(data) {
- this.data = data;
- this.prev = null;
- this.next = null;
- }
-
- // 属性
- this.head = null;
- this.tail = null;
- this.length = 0;
-}
-```
-
-#### 1.2 实现 append()方法
-
-```js
-// 1.向链表尾部追加数据
-DoublyLinkedList.prototype.append = function (data) {
- var newNode = new Node(data);
- // 若是第一次添加，直接将head和tail指向新节点即可
- if (this.length === 0) {
- this.head = newNode;
- this.tail = newNode;
- } else {
- // 让tail与新节点的prev建立连接
- newNode.prev = this.tail;
- // tail的next指向新节点
- this.tail.next = newNode;
- // 改变tail指针
- this.tail = newNode;
- }
-
- this.length += 1;
-};
-```
-
-##### append()方法图解
-
-情况 1：第一次添加 ==> 只需要让 head 和 tail 都指向新节点即可
-
-
- 双向链表结构
-
- 
-
-
-情况 2：不是第一次添加，需要改变 tail 指针的引用关系
-
-- 通过`newNode.prev = this.tail`：建立指向 1
-- 通过`this.tail.next = newNode`：建立指向 2
-
-如图:
-
-
-
- 
-
-
-- 通过`this.tail = newNode`：建立指向 3
-- 要注意**指针改变的顺序**，最后修改 tail 指向，这样未修改前 tail 始终指向原链表的最后一个节点。
-
-如图:
-
-
-
- 
-
-
-#### 1.3 实现 forwardString()、backwardString()、toString()方法
-
-便于我们测试代码，先实现以上三个方法,这里就放在同一个代码块下了～其实 toString()方法就是 forwardString()方法
-
-```js
-// 返回正向遍历节点字符串形式
-DoublyLinkedList.prototype.forwardString = function () {
- var current = this.head;
- var resultString = '';
- while (current) {
- resultString += current.data + ' ';
- current = current.next;
- }
- return resultString;
-};
-
-// 返回反向遍历的节点的字符串形式
-DoublyLinkedList.prototype.backwardString = function () {
- var current = this.tail;
- var resultString = '';
- while (current) {
- resultString += current.data + ' ';
- current = current.prev;
- }
- return resultString;
-};
-
-DoublyLinkedList.prototype.toString = function () {
- return this.forwardString();
-};
-```
-
-##### 测试代码
-
-```js
-// 测试代码
-var dll = new DoublyLinkedList();
-
-dll.append('10');
-dll.append('20');
-dll.append('30');
-console.log(dll.toString()); // 10 20 30
-console.log(dll.forwardString()); // 10 20 30
-console.log(dll.backwardString()); // 30 20 10
-```
-
-#### 1.4 实现 insert()方法
-
-```js
-// 2.向链表中插入数据
-DoublyLinkedList.prototype.insert = function (position, data) {
- // 1.边界判断
- if (position < 0 || position > this.length) return false;
- var newNode = new Node(data);
-
- // 2.第一次添加
- if (this.length === 0) {
- this.head = newNode;
- this.tail = newNode;
- } else {
- if (position === 0) {
- // 3.1 在头部插入
- this.head.prev = newNode;
- newNode.next = this.head;
- this.head = newNode;
- } else if (position === this.length) {
- // 3.2 在尾部插入
- newNode.prev = this.tail;
- this.tail.next = newNode;
- this.tail = newNode;
- } else {
- // 3.3 在中间插入 0 < position < this.length
- var current = this.head;
- var index = 0;
- // 找到要插入的位置
- while (index++ < position) {
- current = current.next;
- }
- // 让新节点的next 指向要插入位置的节点(current)
- newNode.next = current;
- // 让新节点的prev 指向 current的prev
- newNode.prev = current.prev;
- // 将新节点与 当前节点的 前一个节点的 next建立关系
- current.prev.next = newNode;
- // 让newNode成为current的前一个节点
- current.prev = newNode;
- }
- }
- this.length += 1;
- return true;
-};
-```
-
-##### insert 方法图解
-
-我们一起来分析，双向链表在进行插入时，都有哪些情况会出现
-
-1. **情况 1**：当是第一次添加的时候，只需要让 head 和 tail 都指向 newNode 即可
-
-```js
-// 2.第一次添加
-if (this.length === 0) {
- this.head = newNode;
- this.tail = newNode;
-}
-```
-
-
-
- 
-
-
-2. **当不是第一次插入的时候，又可以细分为以下几种情况**
- - 2.1 在头部插入 (position === 0)
- - 2.2 在尾部插入 (position === this.length)
- - 2.3 在中间插入 (0 < position < this.length)
-
-2.1 **在头部插入时**
-
-```js
-if (position === 0) {
- // 3.1 在头部插入
- this.head.prev = newNode;
- newNode.next = this.head;
- this.head = newNode;
-}
-```
-
-- 在 position === 0 时，我们要将 newNode 插入到如图的位置，成为新的头节点
-
-
-
- 
-
-
-- 通过`this.head.prev = newNode`,将头节点的 prev 指向新节点，建立引用 1
-- 通过`newNode.next = this.head`,将新节点的 next 指向原来的头节点，建立引用 2
-- 此时可以发现，newNode 已经与原来的头节点互相建立指向关系
-
-
-
- 
-
-
-- 最后通过` this.head = newNode`,将 newNode 成为新的头节点，建立引用 3
-
-
-
- 
-
-
-2.2 **在尾部插入时**
-
-```js
-if (position === this.length) {
- // 3.2 在尾部插入
- this.tail.next = newNode;
- newNode.prev = this.tail;
- this.tail = newNode;
-}
-```
-
-- 在 position === this.length 时，我们要将 newNode 成为新的尾节点
-
-
-
- 
-
-
-- 通过`this.tail.next = newNode`,将原来的节点的 next 指向新节点，建立引用 1
-- 通过`newNode.prev = this.tail`,将新节点的 prev，指向原来的尾节点，建立引用 2
-
-
-
- 
-
-
-- 通过`this.tail = newNode`,将 tail 指向新的尾节点，建立引用 3
-
-
-
- 
-
-
-2.3 **在中间插入时**
-
-> 这种情况下，我们需要改变的引用关系稍微多一些
-
-```js
-// 3.3 在中间插入 0 < position < this.length
-var current = this.head;
-var index = 0;
-// 找到要插入的位置
-while (index++ < position) {
- current = current.next;
-}
-// 让新节点的next 指向要插入位置的节点(current)
-newNode.next = current;
-// 让新节点的prev 指向 current的prev
-newNode.prev = current.prev;
-// 将新节点与 当前节点的 前一个节点的 next建立关系
-current.prev.next = newNode;
-// 让newNode成为current的前一个节点
-current.prev = newNode;
-```
-
-- 在 0 < position < this.length 时，我们假设在 1 的位置插入一个新的节点，如下图所示
-
-
-
- 
-
-
-- 我们先要找到要插入元素合适的位置，我们可以通过以下代码来找到插入的位置
-
-
-
- 
-
-
-- 我们找到**current 就是新的节点要插入的位置**，所以下面要做的就是让新的节点与 node1，node2**建立起引用关系**
-
-
-
- 
-
-
-- 通过`newNode.next = current`, 让新节点的 next 指向要插入位置的节点(current)，建立引用 1
-- 通过`newNode.prev = current.prev`,让新节点的 prev 指向 current 的 prev，建立引用 2
-
-
-
- 
-
-
-- 通过`current.prev.next = newNode`,将新节点与 当前节点的 前一个节点的 next 建立关系，建立引用 3
-- 通过`current.prev = newNode`,让 newNode 成为 current 的前一个节点，建立引用 4
-
-
-
- 
-
-
-##### 测试代码
-
-```js
-dll.insert(0, '000');
-dll.insert(2, '222');
-dll.insert(5, '555');
-console.log(dll.toString()); // 000 10 222 20 30 555
-```
-
-#### 1.5 实现 get()方法
-
-```js
-// 3.获取对应位置数据
-DoublyLinkedList.prototype.get = function (position) {
- if (position < 0 || position >= this.length) return null;
-
- if (Math.floor(this.length / 2) > position) {
- // 从前往后遍历
- var index = 0;
- var current = this.head;
- while (index++ < position) {
- current = current.next;
- }
- return current.data;
- } else {
- // 从后往前遍历
- var index = this.length - 1;
- var current = this.tail;
- while (index-- > position) {
- current = current.prev;
- }
- return current.data;
- }
-};
-```
-
-##### 测试代码
-
-```js
-console.log(dll.get(0)); // 000
-console.log(dll.get(2)); // 222
-console.log(dll.get(3)); // 20
-console.log(dll.get(5)); // 555
-```
-
-#### 1.6 实现 indexOf()方法
-
-```js
-// 4.获取数据对应的位置
-DoublyLinkedList.prototype.indexOf = function (data) {
- var current = this.head;
- var index = 0;
- while (current) {
- if (current.data === data) return index;
- current = current.next;
- index += 1;
- }
-
- return -1;
-};
-```
-
-##### 测试代码
-
-```js
-console.log(dll.indexOf('000')); // 0
-console.log(dll.indexOf('222')); // 2
-```
-
-#### 1.7 实现 update()方法
-
-其实就在 get 方法找到元素的基础上，赋值新的数据
-
-```js
-// 5.更新方法
-DoublyLinkedList.prototype.update = function (position, newData) {
- if (position < 0 || position >= this.length) return false;
- if (Math.floor(this.length / 2) > position) {
- // 从前往后 遍历
- var current = this.head;
- var index = 0;
- while (index++ < position) {
- current = current.next;
- }
- current.data = newData;
- return true;
- } else {
- // 从后往前 遍历
- var current = this.tail;
- var index = this.length - 1;
- while (index-- > position) {
- current = current.prev;
- }
- current.data = newData;
- return true;
- }
-};
-```
-
-##### 测试代码
-
-```js
-dll.update(1, '111');
-dll.update(3, '333');
-dll.update(4, '444');
-console.log(dll.toString()); // 000 111 222 333 444 555
-```
-
-#### 1.8 实现 removeAt()方法
-
-```js
-// 6.删除指定位置
-DoublyLinkedList.prototype.removeAt = function (position) {
- if (position < 0 || position >= this.length) return null;
- var current = this.head;
-
- // 1.只有一个节点时
- if (this.length === 1) {
- this.head = null;
- this.tail = null;
- } else {
- // 2.删除 头部节点
- if (position === 0) {
- this.head.next.prev = null;
- this.head = this.head.next;
- } else if (position === this.length - 1) {
- // 3.删除尾部节点
- this.tail.prev.next = null;
- this.tail = this.tail.prev;
- // 如果上面一种不好理解，也可以用下面这种写法
- // current = this.tail;
- // this.tail = this.tail.prev;
- // this.tail.next = null;
- } else {
- // 4.删除中间
- var index = 0;
- while (index++ < position) {
- current = current.next;
- }
- current.next.prev = current.prev;
- current.prev.next = current.next;
- }
- }
-
- this.length -= 1;
- return current.data;
-};
-```
-
-##### removeAt()方法图解
-
-让我们一起回顾一下 removeAt 主要实现了什么功能？
-
-removeAt()方法：删除指定位置的元素，删除成功返回删除的元素的数据，删除失败返回 null
-
-下面我们一起分析下，删除时会有哪些情况？
-
-1. **情况 1**:只有一个节点时，将 head 和 tail 指针，指向 null 即可
-
-
-
- 
-
-
-2. 当链表的节点数量大于 1 时，会出现以下三种情况
- - 2.1 删除 头部节点，也就是删除位置 0 上的节点
- - 2.2 删除 尾部节点，也就是链表最后一位的节点
- - 2.3 删除 中间节点
-
-2.1 删除 头部节点
-
-```js
-// 2.删除 头部节点
-if (position === 0) {
- this.head.next.prev = null;
- this.head = this.head.next;
-}
-```
-
-- 如下图所示，我们要删除的节点是 Node1
-
-
-
- 
-
-
-- 通过`this.head.next.prev = null`,让 Node1 后面的节点指向 null(图中引用 1) ，与 Node1 失去引用关系
-- 通过`this.head = this.head.next`,让 head 与 Node1 后面的节点建立引用 2
-- 经过上面两步的操作，Node1 将没有别的节点指向它，最终会被垃圾回收机制给**回收**掉，也就是会被删除
-
-
-
- 
-
-
-2.2 删除尾部节点
-
-```js
-if (position === this.length - 1) {
- // 3.删除尾部节点
- this.tail.prev.next = null;
- this.tail = this.tail.prev;
-}
-```
-
-- 通过`this.tail.prev.next = null`,让尾元素的前一个节点，与之失去引用关系 1
-- 通过`this.tail = this.tail.prev`,让尾元素的前一个节点与 tail 建立引用 2，使之成为新的尾元素
-- 之前的尾元素，由于没有其他元素指向它，最终会被垃圾回收机制删除掉
-
-
-
- 
-
-
-
-```js
-// 4.删除中间
-var index = 0;
-while (index++ < position) {
- current = current.next;
-}
-current.next.prev = current.prev;
-current.prev.next = current.next;
-```
-
-- 先通过 while 循环找到要删除的元素，比如 position = x，那么需要删除的节点就是 Node(x+1)，如下图所示，我们要做的就是断了前后元素的引用关系
-
-
-
-
- 
-
-
-- 通过`current.next.prev = current.prev`,建立新的引用 1
-- 通过`current.prev.next = current.next`,建立新的引用 2
-- current 同样会被垃圾回收掉
-
-
-
- 
-
-
-##### 测试代码
-
-```js
-console.log(dll.removeAt(0)); // 000
-console.log(dll.removeAt(1)); // 222
-console.log(dll.removeAt(3)); // 555
-console.log(dll.toString()); // 111 333 444
-```
-
-#### 1.9 实现 remove()方法
-
-```js
-// 7.remove
-DoublyLinkedList.prototype.remove = function (data) {
- var position = this.indexOf(data);
- return this.removeAt(position);
-};
-```
-
-##### 测试代码
-
-```js
-console.log(dll.remove('111')); // 111
-console.log(dll.toString()); // 333 444Ï
-```
-
-### 2.封装完的双向链表
-
-```js
-function DoublyLinkedList() {
- // 内部类
- function Node(data) {
- this.data = data;
- this.prev = null;
- this.next = null;
- }
-
- // 属性
- this.head = null;
- this.tail = null;
- this.length = 0;
-
- // 1.向链表尾部追加数据
- DoublyLinkedList.prototype.append = function (data) {
- var newNode = new Node(data);
- // 若是第一次添加，直接将head和tail指向新节点即可
- if (this.length === 0) {
- this.head = newNode;
- this.tail = newNode;
- } else {
- // 让tail与新节点的prev建立连接
- newNode.prev = this.tail;
- // tail的next指向新节点
- this.tail.next = newNode;
- // 改变tail指针
- this.tail = newNode;
- }
-
- this.length += 1;
- };
-
- // 2.向链表中插入数据
- DoublyLinkedList.prototype.insert = function (position, data) {
- // 1.边界判断
- if (position < 0 || position > this.length) return false;
- var newNode = new Node(data);
-
- // 2.第一次添加
- if (this.length === 0) {
- this.head = newNode;
- this.tail = newNode;
- } else {
- if (position === 0) {
- // 3.1 在头部插入
- this.head.prev = newNode;
- newNode.next = this.head;
- this.head = newNode;
- } else if (position === this.length) {
- // 3.2 在尾部插入
- this.tail.next = newNode;
- newNode.prev = this.tail;
- this.tail = newNode;
- } else {
- // 3.3 在中间插入 0 < position < this.length
- var current = this.head;
- var index = 0;
- // 找到要插入的位置
- while (index++ < position) {
- current = current.next;
- }
- // 让新节点的next 指向要插入位置的节点(current)
- newNode.next = current;
- // 让新节点的prev 指向 current的prev
- newNode.prev = current.prev;
- // 将新节点与 当前节点的 前一个节点的 next建立关系
- current.prev.next = newNode;
- // 让newNode成为current的前一个节点
- current.prev = newNode;
- }
- }
- this.length += 1;
- return true;
- };
-
- // 3.获取对应位置数据
- DoublyLinkedList.prototype.get = function (position) {
- if (position < 0 || position >= this.length) return null;
-
- if (Math.floor(this.length / 2) > position) {
- // 从前往后遍历
- var index = 0;
- var current = this.head;
- while (index++ < position) {
- current = current.next;
- }
- return current.data;
- } else {
- // 从后往前遍历
- var index = this.length - 1;
- var current = this.tail;
- while (index-- > position) {
- current = current.prev;
- }
- return current.data;
- }
- };
-
- // 4.获取数据对应的位置
- DoublyLinkedList.prototype.indexOf = function (data) {
- var current = this.head;
- var index = 0;
- while (current) {
- if (current.data === data) return index;
- current = current.next;
- index += 1;
- }
-
- return -1;
- };
-
- // 5.更新方法
- DoublyLinkedList.prototype.update = function (position, newData) {
- if (position < 0 || position >= this.length) return false;
- if (Math.floor(this.length / 2) > position) {
- // 从前往后 遍历
- var current = this.head;
- var index = 0;
- while (index++ < position) {
- current = current.next;
- }
- current.data = newData;
- return true;
- } else {
- // 从后往前 遍历
- var current = this.tail;
- var index = this.length - 1;
- while (index-- > position) {
- current = current.prev;
- }
- current.data = newData;
- return true;
- }
- };
-
- // 6.删除指定位置
- DoublyLinkedList.prototype.removeAt = function (position) {
- if (position < 0 || position >= this.length) return null;
- var current = this.head;
-
- // 1.只有一个节点时
- if (this.length === 1) {
- this.head = null;
- this.tail = null;
- } else {
- // 2.删除 头部节点
- if (position === 0) {
- this.head.next.prev = null;
- this.head = this.head.next;
- } else if (position === this.length - 1) {
- // 3.删除尾部节点
- this.tail.prev.next = null;
- this.tail = this.tail.prev;
- // 如果上面一种不好理解，也可以用下面这种写法
- // current = this.tail;
- // this.tail = this.tail.prev;
- // this.tail.next = null;
- } else {
- // 4.删除中间
- var index = 0;
- while (index++ < position) {
- current = current.next;
- }
- current.next.prev = current.prev;
- current.prev.next = current.next;
- }
- }
-
- this.length -= 1;
- return current.data;
- };
-
- // 7.remove
- DoublyLinkedList.prototype.remove = function (data) {
- var position = this.indexOf(data);
- return this.removeAt(position);
- };
-
- DoublyLinkedList.prototype.size = function () {
- return this.length;
- };
-
- DoublyLinkedList.prototype.isEmpty = function () {
- return this.length === 0;
- };
-
- DoublyLinkedList.prototype.toString = function () {
- return this.forwardString();
- };
-
- // 返回正向遍历节点字符串形式
- DoublyLinkedList.prototype.forwardString = function () {
- var current = this.head;
- var resultString = '';
- while (current) {
- resultString += current.data + ' ';
- current = current.next;
- }
- return resultString;
- };
-
- // 返回反向遍历的节点的字符串形式
- DoublyLinkedList.prototype.backwardString = function () {
- var current = this.tail;
- var resultString = '';
- while (current) {
- resultString += current.data + ' ';
- current = current.prev;
- }
- return resultString;
- };
-}
-
-// 测试代码
-var dll = new DoublyLinkedList();
-
-dll.append('10');
-dll.append('20');
-dll.append('30');
-console.log(dll.toString()); // 10 20 30
-console.log(dll.forwardString()); // 10 20 30
-console.log(dll.backwardString()); // 30 20 10
-
-dll.insert(0, '000');
-dll.insert(2, '222');
-dll.insert(5, '555');
-console.log(dll.toString()); // 000 10 222 20 30 555
-
-console.log(dll.get(0)); // 000
-console.log(dll.get(2)); // 222
-console.log(dll.get(3)); // 20
-console.log(dll.get(5)); // 555
-
-console.log(dll.indexOf('000')); // 0
-console.log(dll.indexOf('222')); // 2
-
-dll.update(1, '111');
-dll.update(3, '333');
-dll.update(4, '444');
-console.log(dll.toString()); // 000 111 222 333 444 555
-
-console.log(dll.removeAt(0)); // 000
-console.log(dll.removeAt(1)); // 222
-console.log(dll.removeAt(3)); // 555
-console.log(dll.toString()); // 111 333 444
-
-console.log(dll.remove('111')); // 111
-console.log(dll.toString()); // 333 444
-```
diff --git a/docs/data-structure/05_set.md b/docs/data-structure/05_set.md
deleted file mode 100644
index f295804..0000000
--- a/docs/data-structure/05_set.md
+++ /dev/null
@@ -1,349 +0,0 @@
----
-nav:
- title: 数据结构
- order: 1
-group:
- title: 常见数据结构
- order: 1
-order: 5
----
-
-# 集合(set)
-
-## 一、介绍
-
-集合通常是由一组**无序的**、**不能重复的**元素构成。数学中常指的集合中的元素是可以重复的，但是计算机中集合的元素不能重复。
-
-集合是特殊的数组。
-
-- 特殊之处在于里面的元素**没有顺序**，也**不能重复**。
-- 没有顺序意味着**不能通过下标值进行访问**，不能重复意味着相同的对象在集合中**只会存在一份**。
-
-## 二、集合的实现
-
-集合比较常见的实现方式是哈希表，这里使用 JavaScript 的 Object 进行封装
-
-### 集合常见的操作
-
-- `add(value)` 向集合添加一个新的项
-- `remove(value)` 从集合移除一个值
-- `has(value)` 如果值在集合中，返回 `true`，否则返回` false`
-- `clear()` 移除集合中的所有项
-- `size()` 返回集合所包含元素的数量。与数组的 `length` 属性类似
-- `values()` 返回一个包含集合中所有值的数组。
-- 还有其他的方法，用的不多，这里不做封装
-
-### 1.Set 类的代码实现
-
-```js
-function MySet() {
- this.items = {};
-
- MySet.prototype.add = function (value) {
- if (this.has(value)) return false;
- this.items[value] = value;
- return true;
- };
-
- MySet.prototype.has = function (value) {
- return this.items.hasOwnProperty(value);
- };
-
- MySet.prototype.remove = function (value) {
- if (!this.has(value)) return false;
- delete this.items[value];
- return true;
- };
-
- MySet.prototype.clear = function () {
- this.items = {};
- };
-
- MySet.prototype.size = function () {
- return Object.keys(this.items).length;
- };
-
- // 返回一个包含集合中所有值的数组
- MySet.prototype.values = function () {
- return Object.keys(this.items);
- };
-}
-```
-
-#### 测试代码
-
-```js
-// 测试代码
-var set = new MySet();
-console.log(set.add('10')); // true
-console.log(set.add('10')); // false
-console.log(set.add('20')); // true
-console.log(set.add('30')); // true
-console.log(set.values()); // [ '10', '20', '30' ]
-
-console.log(set.remove('10')); // true
-console.log(set.remove('10')); // false
-console.log(set.values()); // [ '20', '30' ]
-console.log(set.size()); // 2
-```
-
-### 2.集合间操作的封装
-
-- 并集：对于给定的两个集合，返回一个包含两个集合中所有元素的新集合
-- 交集：对于给定的两个集合，返回一个包含两个集合中共有元素的新集合
-- 差集：对于给定的两个集合，返回一个包含所有存在于第一个集合且不存在于第二个集合的元素的新集合
-- 子集：验证一个给定集合是否是另一个集合的子集
-
-
-
- 
-
-
-#### 2.1 并集的实现
-
-```js
-// 并集 ==> 将两个set集合合并
-MySet.prototype.union = function (otherSet) {
- // 创建一个新的set
- var unionSet = new MySet();
- // 取出当前set里的value
- var values = this.values();
-
- // 将当前set里的value添加到新的set中
- for (var i = 0; i < values.length; i++) {
- unionSet.add(values[i]);
- }
-
- // 取出传入的set
- values = otherSet.values();
- // 将传入的set添加到新的set中
- for (var i = 0; i < values.length; i++) {
- // 因为是set结构，所以会过滤掉重复的内容
- unionSet.add(values[i]);
- }
-
- return unionSet;
-};
-```
-
-##### 测试代码
-
-```js
-// 测试代码
-var set1 = new MySet();
-var set2 = new MySet();
-set1.add('10');
-set1.add('20');
-set1.add('30');
-
-set2.add('10');
-set2.add('40');
-set2.add('50');
-var unionSet = set1.union(set2);
-console.log(unionSet.values()); // [ '10', '20', '30', '40', '50' ]
-```
-
-#### 2.2 交集的实现
-
-```js
-// 交集 ==> 取出两个集合重复的内容
-MySet.prototype.intersect = function (otherSet) {
- var intersectSet = new MySet();
- var values = this.values();
-
- for (var i = 0; i < values.length; i++) {
- // 找出重复的内容
- if (otherSet.has(values[i])) {
- // 将重复的内容，添加到交集中
- intersectSet.add(values[i]);
- }
- }
- return intersectSet;
-};
-```
-
-##### 测试代码
-
-```js
-var intersectSet = set1.intersect(set2);
-console.log(intersectSet.values()); // [ '10' ]
-```
-
-#### 2.3 差集的实现
-
-```js
-// 差集 ==> 跟交集相反，剔除集合A中那些在集合B中也存在的内容
-MySet.prototype.difference = function (otherSet) {
- var differenceSet = new MySet();
- var values = this.values();
- for (var i = 0; i < values.length; i++) {
- if (!otherSet.has(values[i])) {
- differenceSet.add(values[i]);
- }
- }
- return differenceSet;
-};
-```
-
-##### 测试代码
-
-```js
-var differenceSet = set1.difference(set2);
-console.log(differenceSet.values()); // [ '20', '30' ]
-```
-
-#### 2.4 判断子集的实现
-
-```js
-// 子集 ==> 若集合B里的内容，在集合A中也存在，那么B就是A的子集
-MySet.prototype.subset = function (otherSet) {
- var values = this.values();
- for (var i = 0; i < values.length; i++) {
- if (!otherSet.has(values[i])) {
- return false;
- }
- }
- return true;
-};
-```
-
-##### 测试代码
-
-```js
-console.log(set1.subset(set2)); // false
-```
-
-### 3.完整代码
-
-```js
-function MySet() {
- this.items = {};
-
- MySet.prototype.add = function (value) {
- if (this.has(value)) return false;
- this.items[value] = value;
- return true;
- };
-
- MySet.prototype.has = function (value) {
- return this.items.hasOwnProperty(value);
- };
-
- MySet.prototype.remove = function (value) {
- if (!this.has(value)) return false;
- delete this.items[value];
- return true;
- };
-
- MySet.prototype.clear = function () {
- this.items = {};
- };
-
- MySet.prototype.size = function () {
- return Object.keys(this.items).length;
- };
-
- // 返回一个包含集合中所有值的数组
- MySet.prototype.values = function () {
- return Object.keys(this.items);
- };
-
- // 并集 ==> 将两个set集合合并
- MySet.prototype.union = function (otherSet) {
- // 创建一个新的set
- var unionSet = new MySet();
- // 取出当前set里的value
- var values = this.values();
-
- // 将当前set里的value添加到新的set中
- for (var i = 0; i < values.length; i++) {
- unionSet.add(values[i]);
- }
-
- // 取出传入的set
- values = otherSet.values();
- // 将传入的set添加到新的set中
- for (var i = 0; i < values.length; i++) {
- // 因为是set结构，所以会过滤掉重复的内容
- unionSet.add(values[i]);
- }
-
- return unionSet;
- };
-
- // 交集 ==> 取出两个集合重复的内容
- MySet.prototype.intersect = function (otherSet) {
- var intersectSet = new MySet();
- var values = this.values();
-
- for (var i = 0; i < values.length; i++) {
- // 找出重复的内容
- if (otherSet.has(values[i])) {
- // 将重复的内容，添加到交集中
- intersectSet.add(values[i]);
- }
- }
- return intersectSet;
- };
-
- // 差集 ==> 跟交集相反，剔除集合A中那些在集合B中也存在的内容
- MySet.prototype.difference = function (otherSet) {
- var differenceSet = new MySet();
- var values = this.values();
- for (var i = 0; i < values.length; i++) {
- if (!otherSet.has(values[i])) {
- differenceSet.add(values[i]);
- }
- }
- return differenceSet;
- };
-
- // 子集 ==> 若集合B里的内容，在集合A中也存在，那么B就是A的子集
- MySet.prototype.subset = function (otherSet) {
- var values = this.values();
- for (var i = 0; i < values.length; i++) {
- if (!otherSet.has(values[i])) {
- return false;
- }
- }
- return true;
- };
-}
-
-// 测试代码
-var set1 = new MySet();
-var set2 = new MySet();
-set1.add('10');
-set1.add('20');
-set1.add('30');
-
-set2.add('10');
-set2.add('40');
-set2.add('50');
-var unionSet = set1.union(set2);
-console.log(unionSet.values()); // [ '10', '20', '30', '40', '50' ]
-
-var intersectSet = set1.intersect(set2);
-console.log(intersectSet.values()); // [ '10' ]
-
-var differenceSet = set1.difference(set2);
-console.log(differenceSet.values()); // [ '20', '30' ]
-
-console.log(set1.subset(set2)); // false
-```
-
-## 三、应用
-
-### 1.[两个数组的交集](https://leetcode-cn.com/problems/intersection-of-two-arrays/)
-
-```js
-/**
- * @param {number[]} nums1
- * @param {number[]} nums2
- * @return {number[]}
- */
-var intersection = function (nums1, nums2) {
- // 使用set结构，一行代码就可以解决
- return [...new Set(nums1)].filter((n) => nums2.includes(n));
-};
-```
diff --git a/docs/data-structure/06_dictionary.md b/docs/data-structure/06_dictionary.md
deleted file mode 100644
index f3e7a16..0000000
--- a/docs/data-structure/06_dictionary.md
+++ /dev/null
@@ -1,207 +0,0 @@
----
-nav:
- title: 数据结构
- order: 1
-group:
- title: 常见数据结构
- order: 1
-order: 6
----
-
-# 字典(Dictionary)
-
-## 一、介绍
-
-字典是用来存储唯一值的一种数据结构，通常以 **[键, 值]** **对 **的形式来存储数据，主要特点是**一一对应**，也称为**映射**。
-
-在 js 中字典是使用`Map`类实现的，每个语言实现的类名不同，比如：Python 中是 `dict`，Java 中的 `HashMap` 和 `TreeMap`，但是实现字典的特点都是一样的
-
-## 二、字典的实现
-
-> 为了区分 ES6 中的 Map，这里使用 Dictionary 作为类名
-
-### 字典常见的操作
-
-- `set(key,value)` 向字典中添加新元素
-- `remove(key)` 通过使用键值来从字典中移除键值对应的数据值
-- `has(key)` 如果某个键值存在于这个字典中，则返回 `true`，反之则返回 `false`
-- `get(key)` 通过键值查找特定的数值并返回
-- `clear()` 将这个字典中的所有元素全部删除
-- `size()` 返回字典所包含元素的数量。与数组的 `length` 属性类似
-- `keys()` 将字典所包含的所有键名以数组形式返回
-- `values()` 将字典所包含的所有数值以数组形式返回
-
-### 代码实现
-
-```js
-function Dictionary() {
- this.item = {};
-
- Dictionary.prototype.has = function (key) {
- return this.item.hasOwnProperty(key);
- };
-
- Dictionary.prototype.set = function (key, value) {
- this.item[key] = value;
- };
-
- Dictionary.prototype.remove = function (key) {
- if (!this.has(key)) return false;
- delete this.item[key];
- return true;
- };
-
- Dictionary.prototype.get = function (key) {
- return this.has(key) ? this.item[key] : undefined;
- };
-
- Dictionary.prototype.clear = function () {
- this.item = {};
- };
-
- Dictionary.prototype.size = function () {
- return this.keys().length;
- };
-
- // 返回一个包含集合中所有值的数组
- Dictionary.prototype.keys = function () {
- return Object.keys(this.item);
- };
-
- Dictionary.prototype.values = function () {
- return Object.values(this.item);
- };
-}
-```
-
-##### 测试代码
-
-```js
-// 测试代码
-let map = new Dictionary();
-map.set('10', 10);
-map.set('20', 20);
-map.set('30', 30);
-console.log('[ map ] >', map); // [ map ] > Dictionary { item: { 10: 10, 20: 20, 30: 30 } }
-
-map.has('10'); // ? true
-
-map.remove('20'); // ? true
-console.log('[ map ] >', map); //[ map ] > Dictionary { item: { 10: 10, 30: 30 } }
-
-map.keys(); // ? [ '10', '30' ]
-
-map.values(); // ? [ 10, 30 ]
-map.size(); // ? 2
-map.clear();
-console.log('[ map ] >', map); // [ map ] > Dictionary { item: {} }
-```
-
-## 三、应用
-
-### 1.[两数之和](/js-logs/n-sum#1两数之和)
-
-```js
-/**
- * @param {number[]} nums
- * @param {number} target
- * @return {number[]}
- */
-var twoSum = function (nums, target) {
- var map = new Map();
-
- for (var i = 0; i < nums.length; i++) {
- var n = nums[i];
- var n2 = target - n;
-
- if (map.has(n2)) {
- return [map.get(n2), i];
- } else {
- map.set(n, i);
- }
- }
-};
-```
-
-### 2.[无重复字符的最长子串](/js-logs/sliding-window#3无重复字符的最长子串)
-
-- 思路
-- 用双指针维护一个"滑动窗口"
-- 不断移动右指针，当遇到重复元素，将左指针移动到重复元素的下一位
-- 遍历过程中，记录窗口的长度，返回最大值
-
-注意
-"abbcdefa"，遇到这种情况，要加判断 ==> 重复元素的位置要大于左指针才有效
-
-```js
-/**
- * @param {string} s
- * @return {number}
- */
-var lengthOfLongestSubstring = function (s) {
- let map = new Map();
- let l = 0,
- res = 0;
-
- for (let r = 0; r < s.length; r++) {
- let c = s[r];
- // 2.如果发现map中已经存过c的位置
- // 4.map.get(c)的位置要大于左指针的位置才更新
- if (map.has(c) && map.get(c) >= l) {
- // 2.1将c的位置拿出来，改变l指针的位置
- l = map.get(c) + 1;
- }
- // 3.更新最长的子串
- res = Math.max(res, r - l + 1);
- // 1.右指针在滑动的过程中，将对应的位置，存入map中
- map.set(c, r);
- }
- return res;
-};
-```
-
-### 3.[有效的括号](/js-logs/stack#20有效的括号)
-
-- 思路：
-- 使用栈结构先进后出的特性
-- 先声明一个栈 stack 变量
-- 遍历整个字符串
-- 发现 ( 、 {、 [ 就入栈
-- 发现 )、}、] 就和栈顶做比较，若栈顶存在对应的括号，将与之对应的括号出栈
-- 遍历结束后，若栈空，则说明所有括号都能找到相匹配的
-- 这题使用 map 做了存储上的优化，不然要写一堆的判断
-
-```js
-/**
- * @param {string} s
- * @return {boolean}
- */
-var isValid = function (s) {
- if (s.length % 2 === 1) {
- return false;
- }
- var stack = [];
- var length = s.length;
- var map = new Map();
- map.set('(', ')');
- map.set('{', '}');
- map.set('[', ']');
-
- for (var i = 0; i < length; i++) {
- var c = s[i];
- if (map.has(c)) {
- stack.push(c);
- } else {
- var stackTop = stack[stack.length - 1];
- if (map.get(stackTop) === c) {
- // 栈顶元素出栈
- stack.pop();
- } else {
- // 栈顶元素与当前元素不匹配 “})” "])" return
- return false;
- }
- }
- }
- return stack.length === 0 ? true : false;
-};
-```
diff --git a/docs/data-structure/07_hash_table.md b/docs/data-structure/07_hash_table.md
deleted file mode 100644
index 28023af..0000000
--- a/docs/data-structure/07_hash_table.md
+++ /dev/null
@@ -1,700 +0,0 @@
----
-nav:
- title: 数据结构
- order: 1
-group:
- title: 常见数据结构
- order: 1
-order: 7
----
-
-# 哈希表(HashTable)
-
-## 一、介绍
-
-哈希表并不好理解，不像数组、链表和树等可通过图形的形式表示其结构和原理。
-
-哈希表的结构就是**数组**，但它神奇之处在于对**下标值的一种变换**，这种变换我们可以称之为**哈希函数**，通过哈希函数可以获取**HashCode**。哈希表通常是基于**数组**实现的，但是相对于数组，它存在更多优势
-
-### 1. 哈希表的优点
-
-- 哈希表可以提供非常快速的**插入-删除-查找操作**；
-- 无论多少数据，插入和删除值都只需要非常短的时间，即 O(1)的时间级。实际上，只需要**几个机器指令**即可完成；
-- 哈希表的速度比**树还要快**，基本可以瞬间查找到想要的元素。
-
-### 2. 哈希表的缺点
-
-- 哈希表中的数据是**没有顺序**的，所以不能以一种固定的方式（比如从小到大 ）来遍历其中的元素。
-- 通常情况下，哈希表中的 key 是**不允许重复**的，不能放置相同的 key，用于保存不同的元素。
-
-**通过以下案例了解哈希表：**
-
-- 案例一：公司想要存储 1000 个人的信息，每一个工号对应一个员工的信息。若**使用数组，增删数据时比较麻烦**；使用**链表，获取数据时比较麻烦**。有没有一种数据结构，能把某一员工的姓名转换为它对应的工号，再根据工号查找该员工的完整信息呢？没错此时就可以使用哈希表的哈希函数来实现。
-- 案例二：存储联系人和对应的电话号码：当要查找张三（比如）的号码时，若使用数组：由于**不知道存储张三数据对象的下标值**，所以**查找起来十分麻烦**，使用**链表时也同样麻烦**。而使用哈希表就能通过**哈希函数**把张三这个名称**转换**为它对应的**下标值**，再通过下标值查找效率就非常高了。
-
-也就是说：哈希表最后还是基于数组来实现的，只不过**哈希表能够通过哈希函数把字符串转化为对应的下标值，建立字符串和下标值的映射关系**。
-
-### 3. 关于哈希表的一些概念
-
-- **哈希化**
- 将**大数字**转化成**数组范围内下标**的过程，称之为哈希化。
-
-- **哈希函数**
- 我们通常会将单词转化成大数字，把大数字进行哈希化的代码实现放在一个函数中，该函数就称为哈希函数。
-
-- **哈希表**
- 对最终数据插入的数组进行整个结构的封装，得到的就是哈希表。
-
-### 4. 认识哈希化
-
-为了把字符串转化为对应的下标值，需要有一套编码系统，为了方便理解我们创建这样一套编码系统：比如 a 为 1，b 为 2，c 为 3，以此类推 z 为 26，空格为 27（不考虑大写情况）。
-
-有了编码系统后，将字母转化为数字也有很多种方案：
-
-- 方案一：数字相加
- 例如 cats 转化为数字：`3 + 1 + 20 + 19 = 43`，那么就把 43 作为 cats 单词的下标值储存在数组中；
- 但是这种方式会存在这样的问题：很多的单词按照该方式转化为数字后都是 43，比如 was。而在数组中一个下标值只能储存一个数据，所以该方式不合理。
-
-- 方案二：幂的连乘
- 我们平时使用的大于 10 的数字，就是用幂的连乘来表示它的唯一性的。 比如： `6543 = 6 * 10^3 + 5 * 10^2 + 4 * 10 + 3`；这样单词也可以用该种方式来表示：`cats = 3 * 27^3 + 1 * 27^2 + 20 * 27 + 17 = 60337`。
- 虽然该方式可以保证字符的唯一性，但是如果是较长的字符（如 aaaaaaaaaa）所表示的数字就非常大，此时要求很大容量的数组，然而其中却有许多下标值指向的是无效的数据（比如不存在 zxcvvv 这样的单词），造成了数组空间的浪费。
-
-两种方案总结：
-
-- 第一种方案（让数字相加求和）产生的数组下标太少。
-- 第二种方案（与 27 的幂相乘求和）产生的数组下标又太多。
-
-现在需要一种压缩方法，把幂的连乘方案系统中得到的**巨大整数范围压缩到可接受的数组范围中**。可以通过==取余==操作来实现。虽然取余操作得到的结构也有可能重复，但是可以通过其他方式解决。
-
-### 5. 解决哈希化后地址仍然冲突的方法
-
-#### 5.1 拉链法
-
-如下图所示，我们将每一个数字都对**10**进行取余操作，则余数的范围**0~9**作为数组的下标值。并且，数组每一个下标值对应的位置存储的不再是一个数字了，而是存储由经过取余操作后得到相同余数的数字组成的**数组**或**链表**
-
-
- 集合间操作
-
- 
-
-
-这样可以根据下标值获取到整个数组或链表，之后继续在数组或链表中查找就可以了。而且，产生冲突的元素一般不会太多。
-
-**总结**：链地址法解决冲突的办法是每个数组单元中存储的不再是单个数据，而是一条链条，这条链条常使用的数据结构为数组或链表，两种数据结构查找的效率相当（因为链条的元素一般不会太多）
-
-#### 5.2 开放地址法
-
-> 开放地址法的主要工作方式是寻找空白的单元格来放置冲突的数据项。
-
-
- 拉链法
-
- 
-
-
-根据探测空白单元格位置方式的不同，可分为三种方法：
-
-- 线性探测
-- 二次探测
-- 再哈希法
-
-##### 5.2.1 线性探测
-
-下面我们根据上图举例感受一下什么是线性探测
-
-- 当插入 13 时：
- 经过哈希化（对 10 取余）之后得到的下标值 index=3，但是该位置已经放置了数据 33。而线性探测就是从 index 位置+1 开始向后一个一个来查找合适的位置来放置 13，所谓合适的位置指的是空的位置，如上图中 index=4 的位置就是合适的位置。
-
-- 当查询 13 时：
- - 首先 13 经过哈希化得到 index=3，如果 index=3 的位置存放的数据与需要查询的数据 13 相同，就直接返回； 不相同时，则线性查找，从 index+1 位置开始一个一个位置地查找数据 13。
- - 查询过程中不会遍历整个哈希表，只要查询到空位置，就停止，因为插入 13 时不会跳过空位置去插入其他位置。
-- 当删除 13 时：
- - 删除操作和上述两种情况类似，但需要注意的是，删除一个数据项时，不能将该位置下标的内容设置为 null，否则会影响到之后其他的查询操作，因为一遇到为 null 的位置就会停止查找。
- - 通常删除一个位置的数据项时，我们可以将它进行特殊处理（比如设置为-1），这样在查找时遇到-1 就知道要继续查找。
-
-**线性探测存在的问题：**
-
-- 线性探测存在一个比较严重的问题，就是**聚集**。
-- 如哈希表中还没插入任何元素时，插入 23、24、25、26、27，这就意味着下标值为 3、4、5、6、7 的位置都放置了数据，这种一连串填充单元就称为聚集。
-- **聚集会影响哈希表的性能，无论是插入/查询/删除都会影响**。
-- 比如插入 13 时就会发现，连续的单元 3~7 都不允许插入数据，并且在插入的过程中需要经历多次这种情况。
-
-
- 开放地址法
-
- 
-
-
-> 二次探测法可以解决该问题
-
-##### 5.2.2 二次探测
-
-**上文所说的线性探测存在的问题：**
-
-- 如果之前的数据是连续插入的，那么新插入的一个数据可能需要探测很长的距离
-
-**二次探测是在线性探测的基础上进行了优化：**
-
-- 线性探测：我们可以看成是步长为 1 的探测，比如从下表值 x 开始，那么线性探测就是按照下标值：x+1、x+2、x+3 等依次探测；
-- 二次探测：对步长进行了优化，比如从下标值 x 开始探测：x+1^2^、x+2^2^、x+3^3^ 。
-- 这样**一次性探测比较长的距离，避免了数据聚集带来的影响**。
-
-**二次探测存在的问题：**
-
-- 当插入数据分布性较大的一组数据时，比如：13-163-63-3-213，这种情况会造成步长不一的一种聚集（虽然这种情况出现的概率较线性探测的聚集要小），同样会影响性能。
-
-##### 5.2.3 再哈希法
-
-在开放地址法中寻找空白单元格的最好的解决方式为再哈希化。
-
-- 二次探测的步长是固定的：1，4，9，16 依次类推
-- 现在需要一种方法：**产生一种依赖关键字(数据)的探测序列**，而不是每个关键字探测步长都一样
-- 这样，**不同的关键字即使映射到相同的数组下标，也可以使用不同的探测序列**
-- 再哈希法的做法为：**把关键字用另一个哈希函数，再做一次哈希化**，用这次**哈希化的结果作为该关键字的步长**
-
-第二次哈希化需要满足以下两点：
-
-- 和第一个哈希函数不同，不然哈希化后的结果仍是原来位置
-- 不能输出为 0，否则每次探测都是原地踏步的死循环
-
-**优秀的哈希函数：**
-
-- stepSize = constant - (key % constant）
-- 其中 constant 是**质数**，且**小于数组的容量**
-- 例如：stepSize = 5 - （key % 5），满足需求，并且结果不可能为 0
-
-**哈希化的效率：**
-
-哈希表中执行插入和搜索操作效率是非常高的。
-
-- 如果没有发生冲突，那么效率就会更高
-- 如果发生冲突，存取时间就依赖后来的探测长度
-- 平均探测长度以及平均存取时间，取决于**填装因子**，\*_随着填装因子变大，探测长度会越来越长。_
-
-**装填因子:**
-
-- 装填因子表示**当前哈希表中已经包含的数据项和整个哈希表长度的比值**
-- 装填因子 = 总数据项 / 哈希表长度
-- 开放地址法的装填因子最大为 1，因为只有空白的单元才能放入元素
-- 链地址法的装填因子可以大于 1，因为只要愿意，拉链法可以无限延伸下去
-
-**不同探测方式性能的比较:**
-
-- 线性探测
-
- 可以看到，随着装填因子的增大，平均探测长度呈指数形式增长，性能较差。实际情况中，最好的装填因子取决于存储效率和速度之间的平衡，随着装填因子变小，存储效率下降，而速度上升。
-
-
- 线性探测
-
- 
-
-
-- 二次探测和再哈希化的性能
-
- 二次探测和再哈希法性能相当，它们的性能比线性探测略好。由下图可知，随着装填因子的变大，平均探测长度呈指数形式增长，需要探测的次数也呈指数形式增长，性能不高。
-
-
-
- 
-
-
-- 链地址法的性能
-
- 可以看到随着装填因子的增加，平均探测长度呈线性增长，较为平缓。在开发中使用链地址法较多，比如 Java 中的 HashMap 中使用的就是链地址法。
-
-
-
- 
-
-
-### 6. 哈希函数
-
-哈希表的优势在于它的速度，所以哈希函数不能采用消耗性能较高的复杂算法。提高速度的一个方法是在哈希函数中尽量减少乘法和除法。
-
-性能高的哈希函数应具备以下两个优点：
-
-- 快速的计算
-- 均匀的分布
-
-#### 快速计算
-
-霍纳法则：在中国霍纳法则也叫做秦久韶算法，具体算法为：
-
-
-
- 
-
-
-求多项式的值时，首先计算最内层括号内一次多项式的值，然后由内向外逐层计算一次多项式的值。这种算法把求 n 次多项式 f(x)的值就转化为求 n 个一次多项式的值。
-
-- 变换之前：
- - 乘法次数：n(n+1) / 2 次
- - 加法次数：n 次
-- 变换之后：
- - 乘法次数：n 次
- - 加法次数：n 次
-
-如果使用大 O 表示时间复杂度的话，直接从变换前的 O(N^2)降到了 O(N)。
-
-#### 均匀分布
-
-在设计哈希表时，我们已经有办法处理映射到相同下标值的情况：链地址法或者开放地址法。但是，为了提供效率，最好的情况还是让数据在哈希表中均匀分布。因此，我们需要在使用常量的地方，尽量使用质数。比如：哈希表的长度、N 次幂的底数等。
-
-**Java 中的 HashMap 采用的是链地址法**，哈希化采用的是公式为：**index = HashCode(key) & (Length-1)** 即将数据化为二进制进行与运算，而不是取余运算。这样计算机直接运算二进制数据，效率更高。但是 JavaScript 在进行较大数据的与运算时会出现问题，所以我们使用 JavaScript 实现哈希化时采用**取余运算**。
-
-## 二、哈希表的封装
-
-### 1. 实现一个哈希函数
-
-```js
-function hashFunction(str, size) {
- // 1.定义hashCode
- var hashCode = 0;
- // 2.霍纳算法，来计算hashCode的值
- for (var i = 0; i < str.length; i++) {
- hashCode = 37 * hashCode + str.charCodeAt(i);
- }
- // 缩小hashCode的范围
- var index = hashCode % size;
- return index;
-}
-```
-
-#### 测试代码
-
-```js
-hashFunction('abc', 7); //? 4
-hashFunction('cba', 7); //? 3
-hashFunction('nba', 7); //? 5
-hashFunction('mba', 7); //? 1
-```
-
-### 2. 哈希表常见操作
-
-- `put(key, value)` 插入或修改操作。
-- `get(key)` 获取哈希表中特定位置的元素。
-- `remove(key)` 删除哈希表中特定位置的元素。
-- `isEmpty()` 如果哈希表中不包含任何元素，返回 `trun`，如果哈希表长度大于 0 则返回 `false`。
-- `size()` 返回哈希表包含的元素个数。
-- `resize(value)` 对哈希表进行扩容操作。
-
-### 3. 创建哈希表类
-
-封装的哈希表的数据结构模型：
-
-
-
- 
-
-
-#### 3.1 创建哈希表类
-
-```js
-// 根据 链地址法（拉链法） 封装哈希表
-function HashTable() {
- this.storage = [];
- this.count = 0; // 保存已经存放的总数，用于计算装载因
- this.limit = 7; // 数组长度，用于动态扩容
-}
-```
-
-#### 3.2 实现 put()方法
-
-哈希表的插入和修改操作是同一个函数：因为，当使用者传入一个 `[key, value]` 时，如果原来不存在该 key，那么就是插入操作，如果原来已经存在该 key，那么就是修改操作。
-
-
-
- 
-
-
-实现思路：
-
-- 首先，根据 key 获取索引值 index，目的为将数据插入到 storage 的对应位置；
-- 然后，根据索引值取出 bucket，如果 bucket 不存在，先创建 bucket，随后放置在该索引值的位置；
-- 接着，判断新增还是修改原来的值。如果已经有值了，就修改该值；如果没有，就执行后续操作。
-- 最后，进行新增数据操作。
-
-**代码实现**
-
-```js
-// 1.插入和修改
-HashTable.prototype.put = function (key, value) {
- if (typeof key === 'String') {
- throw new Error('key must be a string');
- }
-
- // 1.根据key获取对应的index
- var index = this.hashFunction(key, this.limit);
- var bucket = this.storage[index];
- if (bucket === undefined) {
- bucket = [];
- this.storage[index] = bucket;
- }
-
- // 修改操作
- for (var i = 0; i < bucket.length; i++) {
- var tuple = bucket[i];
- if (tuple[0] === key) {
- tuple[1] = value;
- return;
- }
- }
-
- // 插入操作
- bucket.push([key, value]);
- this.count += 1;
-};
-```
-
-为了方便测试，我们下面去实现 get 方法
-
-#### 3.3 实现 get()方法
-
-实现思路：
-
-- 首先，根据 key 通过哈希函数获取它在 `storage` 中对应的索引值 `index`。
-- 然后，根据索引值获取对应的 `bucket`。
-- 接着，判断获取到的 `bucket` 是否为 `null`，如果为 `null`，直接返回 `null`。
-- 随后，线性遍历 `bucket` 中每一个 `key` 是否等于传入的 `key`。如果等于，直接返回对应的 `value`。
-- 最后，遍历完 `bucket` 后，仍然没有找到对应的 `key`，直接 `return null` 即可。
-
-**代码实现**
-
-```js
-// 2.获取操作
-HashTable.prototype.get = function (key) {
- if (typeof key === 'String') {
- throw new Error('key must be a string');
- }
-
- var index = this.hashFunction(key, this.limit);
- var bucket = this.storage[index];
-
- if (bucket === undefined) return null;
- // 从 bucket 中线性查找key
- for (var i = 0; i < bucket.length; i++) {
- var tuple = bucket[i];
- if (tuple[0] === key) {
- return tuple[1];
- }
- }
- // bucket中没有找到返回 null
- return null;
-};
-```
-
-**测试代码**
-
-```js
-//测试代码
-var ht = new HashTable();
-ht.put('abc', '111');
-ht.put('cba', '222');
-ht.put('nba', '333');
-console.log(ht);
-console.log(ht.get('abc')); // ? 111
-console.log(ht.get('cba')); // ? 222
-console.log(ht.get('nba')); // ? 333
-// 测试修改
-ht.put('abc', '444');
-console.log(ht.get('abc')); // ? 444Ï
-```
-
-#### 3.4 实现 remove()方法
-
-实现思路：
-
-- 首先，根据 key 通过哈希函数获取它在 `storage` 中对应的索引值 `index`。
-- 然后，根据索引值获取对应的 `bucket`。
-- 接着，判断获取到的 `bucket` 是否为 `null`，如果为 `null`，直接返回 `null`。
-- 随后，线性查找 `bucket`，寻找对应的数据，并且删除。
-- 最后，依然没有找到，返回 `null`。
-
-```js
-// 3.删除操作
-HashTable.prototype.remove = function (key) {
- if (typeof key === 'String') {
- throw new Error('key must be a string');
- }
-
- var index = this.hashFunction(key, this.limit);
- var bucket = this.storage[index];
-
- if (bucket === undefined) return null;
-
- for (var i = 0; i < bucket.length; i++) {
- var tuple = bucket[i];
- if (tuple[0] === key) {
- bucket.splice(i, 1);
- this.count -= 1;
- return tuple[1];
- }
- }
-
- return null;
-};
-```
-
-#### 3.5 实现 isEmpty()方法和 size()方法
-
-```js
-HashTable.prototype.isEmpty = function () {
- return this.count === 0;
-};
-
-HashTable.prototype.size = function () {
- return this.count;
-};
-```
-
-#### 3.6 哈希表的扩容和压缩
-
-为什么需要扩容？
-
-- 前面我们在哈希表中使用的是长度为 7 的数组，由于使用的是链地址法，装填因子(loadFactor)可以大于 1，所以这个哈希表可以无限制地插入新数据。
-- 但是，随着数据量的增多，storage 中每一个 `index` 对应的 `bucket` 数组（链表）就会越来越长，这就会造成哈希表效率的降低。
-
-什么情况下需要扩容？
-
-- 常见的情况是 `loadFactor > 0.75` 的时候进行扩容。
-
-如何进行扩容？
-
-- 简单的扩容可以直接扩大两倍（关于质数，之后讨论）。
-- 扩容之后所有的数据项都要进行同步修改。
-
-实现思路：
-
-- 首先，定义一个变量，比如 oldStorage 指向原来的 `storage`。
-- 然后，创建一个新的容量更大的数组，让 `this.storage` 指向它。
-- 最后，将 oldStorage 中的每一个 bucket 中的每一个数据取出来依次添加到 `this.storage` 指向的新数组中。
-
-
-
- 
-
-
-#### 3.7 实现 resize()方法
-
-装填因子 = 哈希表中数据 / 哈希表长度，即 `loadFactor = count / HashTable.length`。
-
-resize 方法，既可以实现哈希表的扩容，也可以实现哈希表容量的压缩。
-
-```js
-// 4.哈希表扩容
-HashTable.prototype.resize = function (newLimit) {
- var oldStorage = this.storage;
- this.storage = [];
- this.count = 0;
- this.limit = newLimit;
- for (var i = 0; i < oldStorage.length; i++) {
- var bucket = oldStorage[i];
- if (bucket === undefined) continue;
-
- for (var j = 0; j < bucket.length; j++) {
- var tuple = bucket[j];
- this.put(tuple[0], tuple[1]);
- }
- }
-};
-```
-
-通常情况下当装填因子 `loadFactor > 0.75` 时，对哈希表进行扩容。在哈希表中的 put 方法中添加如下代码，判断是否需要调用扩容函数进行扩容。
-
-```js
-// 动态扩容
-if (this.count > this.limit * 0.75) {
- let newLimit = this.getPrime(this.limit * 2);
- this.resize(newLimit);
-}
-```
-
-当装填因子 `loadFactor < 0.25` 时，对哈希表容量进行压缩。在哈希表中的 remove 方法中添加如下代码，判断是否需要调用扩容函数进行压缩。
-
-```js
-// 动态减少容量
-if (this.limit > 7 && this.count < this.limit * 0.25) {
- let newLimit = this.getPrime(this.limit * 2);
- this.resize(newLimit);
-}
-```
-
-这里封装两个工具函数
-
-```js
-// 判断是否为质数
-HashTable.prototype.isPrime = function (number) {
- var temp = parseInt(Math.sqrt(number));
- for (var i = 2; i <= temp; i++) {
- if (number % i === 0) {
- return false;
- }
- }
- return true;
-};
-
-// 获取一个质数
-HashTable.prototype.getPrime = function (number) {
- while (!this.isPrime(number)) {
- number += 1;
- }
- return number;
-};
-```
-
-## 三、完整实现
-
-```js
-// 根据 链地址法（拉链法） 封装哈希表
-function HashTable() {
- this.storage = [];
- this.count = 0; // 保存已经存放的总数，用于计算装载因
- this.limit = 7; // 数组长度，用于动态扩容
-
- HashTable.prototype.hashFunction = function (str, size) {
- // 1.定义hashCode
- var hashCode = 0;
- // 2.霍纳算法，来计算hashCode的值
- // 这里就采用37作为公式里的质数（无强制要求，质数即可）
- for (var i = 0; i < str.length; i++) {
- hashCode = 37 * hashCode + str.charCodeAt(i);
- }
- // 缩小hashCode的范围
- var index = hashCode % size;
- return index;
- };
-
- // 1.插入和修改
- HashTable.prototype.put = function (key, value) {
- if (typeof key === 'String') {
- throw new Error('key must be a string');
- }
-
- // 1.根据key获取对应的index
- var index = this.hashFunction(key, this.limit);
- var bucket = this.storage[index];
- if (bucket === undefined) {
- bucket = [];
- this.storage[index] = bucket;
- }
-
- // 修改操作
- for (var i = 0; i < bucket.length; i++) {
- var tuple = bucket[i];
- if (tuple[0] === key) {
- tuple[1] = value;
- return;
- }
- }
-
- // 插入操作
- bucket.push([key, value]);
- this.count += 1;
-
- // 动态扩容
- if (this.count > this.limit * 0.75) {
- let newLimit = this.getPrime(this.limit * 2);
- this.resize(newLimit);
- }
- };
-
- // 2.获取操作
- HashTable.prototype.get = function (key) {
- if (typeof key === 'String') {
- throw new Error('key must be a string');
- }
-
- var index = this.hashFunction(key, this.limit);
- var bucket = this.storage[index];
-
- if (bucket === undefined) return null;
- // 从 bucket 中线性查找key
- for (var i = 0; i < bucket.length; i++) {
- var tuple = bucket[i];
- if (tuple[0] === key) {
- return tuple[1];
- }
- }
- // bucket中没有找到返回 null
- return null;
- };
-
- // 3.删除操作
- HashTable.prototype.remove = function (key) {
- if (typeof key === 'String') {
- throw new Error('key must be a string');
- }
-
- var index = this.hashFunction(key, this.limit);
- var bucket = this.storage[index];
-
- if (bucket === undefined) return null;
-
- for (var i = 0; i < bucket.length; i++) {
- var tuple = bucket[i];
- if (tuple[0] === key) {
- bucket.splice(i, 1);
- this.count -= 1;
- // 动态减少容量
- if (this.limit > 7 && this.count < this.limit * 0.25) {
- let newLimit = this.getPrime(this.limit * 2);
- this.resize(newLimit);
- }
- return tuple[1];
- }
- }
-
- return null;
- };
-
- HashTable.prototype.isEmpty = function () {
- return this.count === 0;
- };
-
- HashTable.prototype.size = function () {
- return this.count;
- };
-
- // 4.哈希表扩容
- HashTable.prototype.resize = function (newLimit) {
- var oldStorage = this.storage;
- this.storage = [];
- this.count = 0;
- this.limit = newLimit;
- for (var i = 0; i < oldStorage.length; i++) {
- var bucket = oldStorage[i];
- if (bucket === undefined) continue;
-
- for (var j = 0; j < bucket.length; j++) {
- var tuple = bucket[j];
- this.put(tuple[0], tuple[1]);
- }
- }
- };
-
- // 判断是否为质数
- HashTable.prototype.isPrime = function (number) {
- var temp = parseInt(Math.sqrt(number));
- for (var i = 2; i <= temp; i++) {
- if (number % i === 0) {
- return false;
- }
- }
- return true;
- };
-
- // 获取一个质数
- HashTable.prototype.getPrime = function (number) {
- while (!this.isPrime(number)) {
- number += 1;
- }
- return number;
- };
-}
-
-//测试代码
-
-var ht = new HashTable();
-ht.put('abc', '111');
-ht.put('cba', '222');
-ht.put('nba', '333');
-console.log(ht);
-console.log(ht.get('abc')); // ? 111
-console.log(ht.get('cba')); // ? 222
-console.log(ht.get('nba')); // ? 333
-
-ht.put('abc', '444');
-console.log(ht.get('abc')); // ? 444
-ht.remove('abc');
-console.log(ht.get('abc')); // ? null
-```
diff --git a/docs/data-structure/08_tree&binary_tree.md b/docs/data-structure/08_tree&binary_tree.md
deleted file mode 100644
index b78911e..0000000
--- a/docs/data-structure/08_tree&binary_tree.md
+++ /dev/null
@@ -1,252 +0,0 @@
----
-nav:
- title: 数据结构
- order: 1
-group:
- title: 常见数据结构
- order: 1
-order: 8
----
-
-# 树(Tree)和二叉树(BinaryTree)
-
-## 一、树结构的介绍
-
-我们大家都见过现实生活中的树 🌲，例如：
-
-
-
- 
-
-
-### 1.1 树的特点：
-
-- 树一般都有一个**根**，连接着根的是**树干**
-- 树干会发生**分叉**，形成许多树枝，树枝会继续分化成更小的树枝
-- 树枝的最后是**叶子**
-
-现实生活中很多结构都是树的抽象，模拟的树结构相当于旋转`180°`的树，如下：
-
-
- 世界十大名树全球最著名的树盘点其中一棵位于我国台湾
-
- 
-
-
-### 1.2 树结构 对比于数组/链表/哈希表 有哪些**优势**呢？
-
-**数组：**
-
-- 优点：可以通过**下标值访问**，效率高；
-- 缺点：查找数据时需要先对数据进行**排序**，生成**有序数组**，才能提高查找效率；并且在插入和删除元素时，需要大量的**位移操作**；
-
-**链表：**
-
-- 优点：数据的插入和删除操作效率都很高；
-- 缺点：**查找**效率低，需要从头开始依次查找，直到找到目标数据为止；当需要在链表中间位置插入或删除数据时，插入或删除的效率都不高。
-
-**哈希表：**
-
-- 优点：哈希表的插入/查询/删除效率都非常高；
-- 缺点：**空间利用率不高**，底层使用的数组中很多单元没有被利用；并且哈希表中的元素是**无序**的，不能按照固定顺序遍历哈希表中的元素；而且不能快速找出哈希表中**最大值或最小值**这些特殊值。
-
-**树结构：**
-
-优点：树结构综合了上述三种结构的优点，同时也弥补了它们存在的缺点（虽然效率不一定都比它们高），比如树结构中数据都是有序的，查找效率高；空间利用率高；并且可以快速获取最大值和最小值等。
-
-总结：**每种数据结构都有自己特定的应用场景**
-
-## 二、认识树结构
-
-> **树（Tree）**:由 n（n ≥ 0）个节点构成的**有限集合**。当 n = 0 时，称为**空树**。
-
-对于任一棵非空树（n > 0），它具备以下性质：
-
-- 数中有一个称为 **根（Root）** 的特殊节点，用 **r** 表示；
-- 其余节点可分为 m（m > 0）个互不相交的有限集合 T1，T2，...，Tm，其中每个集合本身又是一棵树，称为原来树的**子树（SubTree）**。
-
-### 2.1 树的常用术语：
-
-
-
- 
-
-
-- **节点的度（Degree）**：节点的**子树个数**，比如节点 B 的度为 2
-- **树的度**：树的所有节点中**最大的度数**，如上图树的度为 2
-- **叶节点（Leaf）**：**度为 0 的节点**（也称为叶子节点），如上图的 H，I，J 等
-- **父节点（Parent）**：度不为 0 的节点称为父节点，如上图节点 B 是节点 D 和 E 的父节点
-- **子节点（Child）**：若 B 是 D 的父节点，那么 D 就是 B 的子节点
-- **兄弟节点（Sibling）**：具有同一父节点的各节点彼此是兄弟节点，比如上图的 B 和 C，D 和 E 互为兄弟节点
-- **路径和路径长度**：路径指的是一个节点到另一节点的通道，路径所包含边的个数称为路径长度，比如 A->H 的路径长度为 3
-- **节点的层次（Level）**：规定**根节点在 1 层**，其他任一节点的层数是其父节点的**层数加 1**。如 B 和 C 节点的层次为 2
-- **树的深度（Depth）**：树种所有节点中的**最大层次**是这棵树的深度，如上图树的深度为 4
-
-### 2.2 数结构的表示方式
-
-#### 2.2.1 常见的表示方法
-
-
-
- 
-
-
-如图，树结构的组成方式类**似于链表**，都是由一个个节点连接构成。
-
-不过，根据每个父节点子节点数量的不同，每一个父节点需要的引用数量也不同。
-
-比如：节点 A 需要 3 个引用，分别指向子节点 B，C，D；而 B 节点需要 2 个引用，分别指向子节点 E 和 F；K 节点由于没有子节点，所以不需要引用。
-
-这种方法**缺点**在于我们**无法确定某一结点的引用数**
-
-#### 2.2.2 **儿子-兄弟表示法**
-
-
-
- 
-
-
-这种表示方法可以完整地记录每个节点的数据，比如
-
-```js
-//节点A
-Node{
- //存储数据
- this.data = data
- //统一只记录左边的子节点
- this.leftChild = B
- //统一只记录右边的第一个兄弟节点
- this.rightSibling = null
-}
-
-//节点B
-Node{
- this.data = data
- this.leftChild = E
- this.rightSibling = C
-}
-
-//节点F
-Node{
- this.data = data
- this.leftChild = null
- this.rightSibling = null
-}
-```
-
-这种表示法的优点在于每一个节点中引用的数量都是确定的
-
-#### 2.2.3 将儿子-兄弟表示法 选择
-
-将儿子-兄弟表示法旋转后会打开一个新世界，让我们一步一步揭开新世界的大门，下面是儿子-兄弟表示法**尚未旋转**的结构
-
-
-
- 
-
-
-将其顺时针旋转 45° 之后：
-
-
-
-这样就成为了一棵**二叉树**，由此我们可以得出结论：**任何树都可以通过二叉树进行模拟**。
-
-问：但是这样父节点不是变了吗？
-
-其实，父节点的设置只是为了方便指向子节点，在代码实现中谁是父节点并没有关系，只要能正确找到对应节点即可
-
-## 三、二叉树的介绍
-
-#### 3.1 什么是二叉树？
-
-
-
- 
-
-
-正如上图所示：如果树中的每一个节点最多只能由**两个子节点**，这样的树就称为**二叉树**
-
-二叉树**十分重要**，不仅仅是因为简单，更是因为几乎所有的树都可以表示成二叉树形式。
-
-#### 3.2 二叉树的组成：
-
-- 二叉树可以为空，也就是可以没有节点；
-- 若二叉树不为空，则它由根节点和称为其**左子树 TL**和**右子树 TR**的两个不相交的二叉树组成；
-
-#### 3.3 二叉树的五种形态
-
-
-
- 
-
-
-上图分别表示：**空的二叉树**、只有**一个节点的二叉树**、只有**左子树 TL**的二叉树、只有**右子树 TR**的二叉树、有左右**两个子树的二叉树**
-
-#### 3.4 二叉树的特性
-
-- 一个二叉树的**第 i 层的最多有 2 ^ (i-1)个节点，i >= 1**
-- 一个**深度为 k 的二叉树的最多有(2 ^ k) - 1 个节点，k >= 1**
-- 对任何非空二叉树，若 n0 表示叶子节点的个数，n2 表示度为 2 的非叶子节点个数，那么两者满足关系：n0 = n2 + 1；
- - 如下图所示：H，E，I，J，G 为叶子节点，总数为 5；
- - A，B，C，F 为度为 2 的非叶子节点，总数为 4；
- - 满足 n0 = n2 + 1 的规律
-
-
-
- 
-
-
-#### 3.5 特殊的二叉树
-
-##### 3.5.1 完美二叉树 ｜满二叉树
-
-完美二叉树（Perfect Binary Tree）也成为满二叉树（Full Binary Tree），在二叉树中，除了最下一层的叶子节点外，每层节点都有 2 个子节点，这就构成了完美二叉树
-
-
-
- 
-
-
-##### 3.5.2 完全二叉树
-
-- 除了**二叉树最后一层外**，其他各层的节点数都达到了**最大值**
-- 并且，最后一层的叶子节点**从左向右是连续存在**，**只缺失右侧若干叶子节点**
-- **完美二叉树是特殊的完全二叉树**
-
-
-
- 
-
-
-在上图中，由于**H 缺失了右子节点**，所以它**不是**完全二叉树，若 H 存在右子节点，那么上图就是一颗完全二叉树
-
-#### 3.6 二叉树的数据存储
-
-常见的二叉树存储方式为**数组和链表**：
-
-##### 3.6.1 使用数组存储数据
-
-**完全二叉树**：按从上到下，从左到右的方式存储数据
-
-
-
- 
-
-
-| 节点 | A | B | C | D | E | F | G | H |
-| :------: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
-| **序号** | **1** | **2** | **3** | **4** | **5** | **6** | **7** | **8** |
-
-使用数组存储时，取数据的时候也十分方便：
-
-- **左子节点的序号等于父节点序号 \* 2**，**右子节点的序号等于父节点序号 \* 2 + 1**
-
-- 例如：H 节点的数据是 D 节点数据 * 2 ，也就是 4*2= 8，I 节点的数据是 D 节点数据 * 2 + 1，也就是 4*2+1 = 9
-
-**非完全二叉树**：非完全二叉树需要转换成完全二叉树才能按照上面的方案存储，这样会**浪费**很大的**存储空间**
-
-
-
- 
-
-
-| 节点 | A | B | C | ^ | ^ | F | ^ | ^ | ^ | ^ | ^ | ^ | M |
-| :------: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :----: | :----: | :----: | :----: |
-| **序号** | **1** | **2** | **3** | **4** | **5** | **6** | **7** | **8** | **9** | **10** | **11** | **12** | **13** |
-
-所以，对于非完全二叉树，一般都是使用**链表**进行数据存储
-
-##### 3.6.2 使用链表存储数据
-
-每一个节点封装成一个 Node，Node 中包含存储的数据、左节点的引用和右节点的引用
-
-
-
- 
-
diff --git a/docs/data-structure/09_binary_search_tree.md b/docs/data-structure/09_binary_search_tree.md
deleted file mode 100644
index 1e834db..0000000
--- a/docs/data-structure/09_binary_search_tree.md
+++ /dev/null
@@ -1,1308 +0,0 @@
----
-nav:
- title: 数据结构
- order: 1
-group:
- title: 常见数据结构
- order: 1
-order: 9
----
-
-# 二叉搜索树(BinarySearchTree)
-
-## 一、对二叉搜索树的介绍
-
-> 二叉搜索树（BST，Binary Search Tree），也称为二叉排序树和二叉查找树
-
-### 1. 二叉搜索树的特性
-
-- 非空左子树的所有键值小于其根节点的键值。比如：下图三，中节点 6 的所有非空左子树的键值都小于 6
-- 非空右子树的所有键值大于其根节点的键值；比如：下图三，中节点 6 的所有非空右子树的键值都大于 6；
-- 左、右子树本身也都是二叉搜索树
-
-
-
- 
-
-
-图二和图三符合二叉搜索树的特效，所以属于二叉树
-
-图一中的节点 5 要比节点 7 大，不符合二叉搜索树的特性， 所以不是二叉树
-
-**总结**：二叉搜索树的特点主要是**较小的值总是保存在左节点上**，相对**较大的值总是保存在右节点**上。这种特点使得二叉搜索树的**查询效率非常高**，这也就是二叉搜索树中“搜索”的来源。
-
-### 2. 二叉搜索树应用举例
-
-
-
- 
-
-
-上面是一个二叉搜索树，我们若想在其中找到数据 10，只需要查找 4 次，查找效率非常高有木有！
-
-查找步骤图解：
-
-
-
- 
-
-
-- 第 1 次：将 10 与根节点 9 进行比较，由于 10 > 9，所以 10 下一步与根节点 9 的右子节点 13 比较；
-- 第 2 次：由于 10 < 13，所以 10 下一步与父节点 13 的左子节点 11 比较；
-- 第 3 次：由于 10 < 11，所以 10 下一步与父节点 11 的左子节点 10 比较；
-- 第 4 次：由于 10 = 10，最终查找到数据 10
-- 总结：有没有觉得很像**二分查找**
-
-下面看一下数组这种数据结构 的查找效率：
-
-
-
- 
-
-
-同样是 15 个数据，在排序好的数组中查询数据 10，需要查询 10 次
-
-其实：如果是排序好的数组，可以通过二分查找：第一次找 9，第二次找 13，第三次找 15...。我们发现如果把**每次二分的数据拿出来以树的形式表示的话就是二叉搜索树**。这就是数组二分法查找效率之所以高的原因
-
-### 3. 二叉搜索树的基本属性
-
-
-
- 
-
-
-二叉搜索树有四个最基本的属性：指向节点的**根**（root），节点中的**键**（key）、**左指针**（right）、**右指针**（right）
-
-用代码的方式体现：
-
-```js
-function BinarySearchTree() {
- // 节点类
- function Node(key) {
- this.left = null; // 左指针
- this.key = key;
- this.right = null; // 右指针
- }
-
- this.root = null; // 根节点
-}
-```
-
-## 二、对二叉搜索树的封装
-
-### 二叉搜索树的常见操作：
-
-- `insert（key）`：向树中插入一个新的键
-- `search（key）`：在树中查找一个键，如果节点存在，则返回 true；如果不存在，则返回 false
-- `midOrderTraversal`：通过中序遍历方式遍历所有节点
-- `preOrderTraversal`：通过先序遍历方式遍历所有节点
-- `postOrderTraversal`：通过后序遍历方式遍历所有节点
-- `min`：返回树中最小的值/键
-- `max`：返回树中最大的值/键
-- `remove（key）`：从树中移除某个键
-
-### 1. 实现 insert()方法
-
-```js
-BinarySearchTree.prototype.insert = function (key) {
- // 1. 创建一个节点类
- var newNode = new Node(key);
- // 2. 判断根节点是否为空
- if (this.root === null) {
- // 2.1 直接插入新的节点
- this.root = newNode;
- } else {
- // 2.2 插入节点类，实现根节点非空时的插入
- this._insertNode(this.root, newNode);
- }
-};
-```
-
-#### 1.1 \_insertNode 方法的实现
-
-思路：比较 传入的两个节点(root 和要插入的节点) 的大小，寻找新节点适合插入的位置，直到成功插入新节点为止
-
-这里先看一眼**代码实现**，然后再看图解步骤：
-
-```js
-BinarySearchTree.prototype._insertNode = function (node, newNode) {
- // 向左寻找
- if (node.key > newNode.key) {
- if (node.left === null) {
- node.left = newNode;
- } else {
- this._insertNode(node.left, newNode);
- }
- } else {
- // 向右寻找
- if (node.right === null) {
- node.right = newNode;
- } else {
- this._insertNode(node.right, newNode);
- }
- }
-};
-```
-
-根据传入两个节点的比较分为两种情况：
-
-##### 1.1.1 newNode.key < node.key 时
-
-根据二叉搜索树的特性，新节点的位置肯定在 node 节点的左子树，
-
-- 情况 1：若 node 节点的左子树是 null，那么说明 node 没有左子树，其实这个时候 newNode 已经找到合适的位置，让 newNode 插入到 node 的 left 即可
-- 情况 2：若 node 节点的左子树**有值存在**，那么需要在 node 的左子树去寻找新节点的插入位置，所以这里需要递归的查找下去并插入面，查找过程如下图所示：
-
-
-
- 
-
-
-##### 1.1.2 newNode.key > node.key 时
-
-这时候遇到的情况，根上面刚好相反，新节点的位置肯定在 node 节点的右子树嘛，也是需要递归的找到合适的插入位置并插入即可，查找插入过程如图所示：
-
-
-
- 
-
-
-#### 测试代码
-
-```js
-// 测试代码
-var bst = new BinarySearchTree();
-
-bst.insert(11);
-bst.insert(7);
-bst.insert(15);
-bst.insert(5);
-bst.insert(9);
-console.log('[ bst ] >', bst);
-```
-
-结构化一下测试数据应该是这样的一颗树：
-
-
-
- 
-
-
-### 2. 对二叉搜索树的遍历
-
-这里所说的树的遍历**不仅仅针对二叉搜索树**，而是**适用于所有的二叉树**。
-
-由于树结构不是线性结构，所以遍历方式有多种选择，常见的三种二叉树遍历方式为：
-
-- 先序遍历 > ==前序位置的代码在刚刚进入一个二叉树节点的时候执行==
-- 中序遍历 > ==后序位置的代码在将要离开一个二叉树节点的时候执行==
-- 后序遍历 > ==中序位置的代码在一个二叉树节点左子树都遍历完，即将开始遍历右子树的时候执行==
-
-还有层序遍历，使用较少，理解上面高亮部分的内容，对刷 leetcode 非常有好处
-
-
-
- 
-
-
-#### 2.1 先序遍历
-
-先序遍历的过程为：
-
-- 首先，遍历根节点
-- 然后，遍历其左子树
-- 最后，遍历其右子树
-
-示例：
-
-
-
- 
-
-
-如上图所示，二叉树的节点遍历顺序为：A -> B -> D -> H -> I -> E -> C -> F -> G。
-
-##### 2.1.1 代码实现
-
-```js
-//1.先序遍历 ==> 根 -> 左 -> 右
-BinarySearchTree.prototype.preOrderTraversal = function (handler) {
- this._preOrderTraversalNode(this.root, handler);
-};
-
-BinarySearchTree.prototype._preOrderTraversalNode = function (node, handler) {
- if (node !== null) {
- // 1.处理经过的节点
- handler(node.key);
- // 2.处理左节点
- this._preOrderTraversalNode(node.left, handler);
- // 3.处理右节点
- this._preOrderTraversalNode(node.right, handler);
- }
-};
-```
-
-##### 2.1.2 过程详解
-
-以遍历以下二叉搜索树为例：
-
-
-
- 
-
-
-首先调用 preOrderTraversal 方法，在方法里再调用 preOrderTraversalNode 方法用于遍历二叉搜索树。在 preOrderTraversalNode 方法中，递归 1 负责遍历左子节点，递归 2 负责遍历右子节点。先执行递归 1，执行过程如下图所示：
-
-> 这里用：preOrderTraversalNode() 表示为 A()，方便图解
-
-
-
- 
-
-
-可以看到一共递归调用了 4 次方法 A，分别传入 11、7、5、3，最后遇到 null 不满足 node != null 条件结束递归 1。
-
-注意此时只是执行完最开始的递归 1，并没有执行递归 2，并且递归 1 执行到 null 停止后要一层层地往上返回，按顺序将调用的函数压出函数调用栈。
-
-关于函数调用栈：之前的四次递归共把 4 个函数压入了函数调用栈，现在递归执行完了一层层地把函数压出栈。
-
-值得注意的是：每一层函数都只是执行完了递归 1，当返回到该层函数时，比如 A（3）要继续执行递归 2 遍历二叉搜索树中的右子节点。
-
-在执行递归 2 的过程中会不断调用方法 A，并依次执行递归 1 和递归 2，以此类推直到遇到 null 不满足 node != null 条件为止，才停止递归并一层层返回，如此循环。同理 A（5）层、A（7）层、A（11）层都要经历上述循环，直到将二叉搜索树中的节点全部遍历完为止。
-
-具体过程如下图所示：
-
-
-
- 
-
-
-##### 2.1.3 测试代码
-
-```js
-// 测试代码
-var bst = new BinarySearchTree();
-
-bst.insert(11);
-bst.insert(7);
-bst.insert(15);
-bst.insert(5);
-bst.insert(3);
-bst.insert(9);
-bst.insert(8);
-bst.insert(10);
-bst.insert(13);
-bst.insert(12);
-bst.insert(14);
-bst.insert(20);
-bst.insert(18);
-bst.insert(25);
-bst.insert(6);
-
-var resString = ' ';
-bst.preOrderTraversal(function (key) {
- resString += key + ' ';
-});
-console.log(resString); // ? 11 7 5 3 6 9 8 10 15 13 12 14 20 18 25
-```
-
-#### 2.2 中序遍历
-
-实现思路：与先序遍历原理相同，只不过是遍历的顺序不一样了。
-
-- 首先，遍历其左子树
-- 然后，遍历根（父）节点
-- 最后，遍历其右子树
-
-##### 2.2.1 代码实现
-
-```js
-// 2.中序遍历 ==> 左 -> 根 -> 右
-BinarySearchTree.prototype.midOrderTraversal = function (handler) {
- this._midOrderTraversalNode(this.root, handler);
-};
-
-BinarySearchTree.prototype._midOrderTraversalNode = function (node, handler) {
- if (node !== null) {
- // 1.处理左节点
- this._midOrderTraversalNode(node.left, handler);
- // 2.处理经过的节点
- handler(node.key);
- // 3.处理右节点
- this._midOrderTraversalNode(node.right, handler);
- }
-};
-```
-
-##### 2.2.2 过程详解
-
-遍历的顺序应如下图所示：
-
-
-
- 
-
-
-首先调用 midOrderTraversal 方法，在方法里再调用 midOrderTraversalNode 方法用于遍历二叉搜索树。
-
-先使用递归 1 遍历左子树中的节点；然后，处理父节点；最后，遍历右子树中的节点。
-
-##### 2.2.3 测试代码
-
-```js
-// 插入顺序和先序遍历的插入一样，这里就省略了，也可以看上图，验证中序遍历的结果
-var resString = ' ';
-bst.midOrderTraversal(function (key) {
- resString += key + ' ';
-});
-console.log(resString); // ? 3 5 6 7 8 9 10 11 12 13 14 15 18 20 25
-```
-
-#### 2.3 后序遍历
-
-实现思路：与先序遍历原理相同，只不过是遍历的顺序不一样了。
-
-- 首先，遍历其左子树
-- 然后，遍历其右子树
-- 最后，遍历根（父）节点
-
-##### 2.3.1 代码实现
-
-```js
-// 3.后序遍历 ==> 左 -> 右 -> 根
-BinarySearchTree.prototype.postOrderTraversal = function (handler) {
- this._postOrderTraversalNode(this.root, handler);
-};
-
-BinarySearchTree.prototype._postOrderTraversalNode = function (node, handler) {
- if (node !== null) {
- // 1.处理左节点
- this._postOrderTraversalNode(node.left, handler);
- // 2.处理右节点
- this._postOrderTraversalNode(node.right, handler);
- // 3.处理经过的节点
- handler(node.key);
- }
-};
-```
-
-##### 2.3.2 过程详解
-
-遍历的顺序应如下图所示：
-
-
-
- 
-
-
-首先调用 postOrderTraversal 方法，在方法里再调用 postOrderTraversalNode 方法用于遍历二叉搜索树。
-
-先使用递归 1 遍历左子树中的节点；然后，遍历右子树中的节点；最后，处理父节点。
-
-##### 2.3.3 测试代码
-
-```js
-var resString = ' ';
-bst.postOrderTraversal(function (key) {
- resString += key + ' ';
-});
-console.log(resString); // ? 3 6 5 8 10 9 7 12 14 13 18 25 20 15 11
-```
-
-### 3. 对二叉搜索树的搜索
-
-#### 3.1 搜索最大值&最小值
-
-在二叉搜索树中查找最值非常简单，最小值在二叉搜索树的最左边，最大值在二叉搜索树的最右边。只需要一直向左/右查找就能得到最值，如下图所示：
-
-
-
- 
-
-
-##### 代码实现
-
-```js
-BinarySearchTree.prototype.min = function () {
- var node = this.root;
- while (node.left !== null) {
- node = node.left;
- }
- return node.key;
-};
-
-BinarySearchTree.prototype.max = function () {
- var node = this.root;
- while (node.right !== null) {
- node = node.right;
- }
- return node.key;
-};
-```
-
-#### 3.2 实现 search()
-
-search 方法可以返回某个值是否在二叉树中，若存在返回 true，不存在返回 false
-
-查找二叉搜索树当中的特定值效率也非常高。只需要从根节点开始将需要查找节点的 key 值与之比较，若**node.key < root**则向左查找，若**node.key > root**就向右查找，直到找到或查找到 null 为止。这里可以使用递归实现，也可以采用循环来实现
-
-##### 3.2.1 代码实现
-
-```js
-BinarySearchTree.prototype.search = function (key) {
- return this._searchNode(this.root, key);
-};
-BinarySearchTree.prototype._searchNode = function (node, key) {
- // 这里使用while循环
- var node = this.root;
- while (node !== null) {
- if (key < node.key) {
- node = node.left;
- } else if (key > node.key) {
- node = node.right;
- } else {
- return true;
- }
- }
- return false;
-
- // 递归调用，占用空间，
- // if (node === null) return false;
- // if (node.key < key) {
- // // 查找右边的树
- // this._searchNode(node.right, key);
- // } else if (node.key > key) {
- // //查找左边的树
- // this._searchNode(node.left, key);
- // } else {
- // // 相等情况，说明找到
- // return true;
- // }
-};
-```
-
-##### 3.2.2 测试代码
-
-```js
-// 测试代码
-var bst = new BinarySearchTree();
-
-bst.insert(11);
-bst.insert(7);
-bst.insert(15);
-bst.insert(5);
-bst.insert(3);
-bst.insert(9);
-bst.insert(8);
-bst.insert(10);
-bst.insert(13);
-bst.insert(12);
-bst.insert(14);
-bst.insert(20);
-bst.insert(18);
-bst.insert(25);
-bst.insert(6);
-
-bst.search(3); // ? true
-bst.search(26124); // ? false
-```
-
-### 4. 实现 remove()方法
-
-> 删除方法要考虑的情况比较多，是一个难点
-
-删除思路：
-
-- 先寻找要删除的节点
- - 若没有找到要删除的节点，说明节点不存在，直接返回 return false 即可
- - 若找到要删除的节点，我们需要先记录下**节点(current)**，以及**删除节点的父节点(parent)**，删除节点是**父节点的左子树还是父节点的右子树(isLeftChild)**
- - 若找到删除的节点，我们需要考虑以下 3 种情况
- - 要删除的是**叶子**节点
- - 要删除的节点，只有**一个子节点**
- - 要删除的节点，有**两个子节点**
-
-#### 4.1 寻找要删除的节点
-
-```js
-BinarySearchTree.prototype.remove = function (key) {
- var current = this.root; // current保存要删除的节点
- var parent = null; // 保存要删除节点的父节点
- var isLeftChild = false; // 用于判断要删除的节点是父节点的左节点还是右节点
-
- // 寻找要删除的节点
- while (current.key != key) {
- parent = current;
- if (key < current.key) {
- isLeftChild = true;
- current = current.left;
- } else {
- isLeftChild = false;
- current = current.right;
- }
- // 若没有找到，说明不需要删除，返回false
- if (current === null) return false;
- }
-};
-```
-
-#### 4.2 找到了要删除的节点
-
-当我们通过上面的代码找到了删除的节点，我们需要对以下 3 种情况进行展开
-
-- 4.2.1 要删除的是**叶子**节点
-- 4.2.2 要删除的节点，只有**一个子节点**
-- 4.2.3 要删除的节点，有**两个子节点**
-
-##### 4.2.1 要删除的是叶子节点
-
-```js
-// 找到了对应的节点,进行情况分析
-// 1.删除的是叶子结点
-if (current.left === null && current.right === null) {
- // 1.1 删除的是根节点
- if (current === this.root) {
- this.root = null;
- } else if (isLeftChild) {
- parent.left = null;
- } else {
- parent.right = null;
- }
-}
-```
-
-###### 情况分析：
-
-当该叶子节点为根节点时，如下图所示，此时**current == this.root**，直接通过：**this.root = null**，删除根节点。
-
-
-
- 
-
-
-当该叶子节点不为根节点时也有两种情况，如下图所示：
-
-
-
- 
-
-
-若 current = 8，可以通过：`parent.left = null`，删除节点 8
-
-若 current = 10，可以通过：`parent.right = null`，删除节点 10
-
-##### 4.2.2 要删除的节点只有一个子节点
-
-```js
-// 此处是要删除的是叶子节点代码...
-
-// 2.删除的节点只有一个子节点
-else if (current.left === null) {
- // 2.1 要删除的是右节点，下面按照一个节点的情况删除即可
- if (current === this.root) {
- this.root = current.right;
- } else if (isLeftChild) {
- parent.left = current.right;
- } else {
- parent.right = current.right;
- }
-} else if (current.right === null) {
- // 2.2 要删除的是左节点
- if (current === this.root) {
- this.root = current.left;
- } else if (isLeftChild) {
- parent.left = current.left;
- } else {
- parent.right = current.left;
- }
-}
-```
-
-###### 情况分析：
-
-当 current 存在左子节点时`current.right === null`：
-
-- 情况 1：current 为根节点`current === this.root`，如节点 11，此时通过：`this.root = current.left`，删除根节点 11
-- 情况 2：current 为父节点 parent 的左子节点`isLeftChild === true`，如节点 5，此时通过：`parent.left = current.left`，删除节点 5
-- 情况 3：current 为父节点 parent 的右子节点`isLeftChild == false`，如节点 9，此时通过：`parent.right = current.left`，删除节点 9
-
-
-
- 
-
-
-当 current 存在右子节点时`current.left === null`：
-
-- 情况 4：current 为根节点`current === this.root`，如节点 11，此时通过：`this.root = current.right`，删除根节点 11
-- 情况 5：current 为父节点 parent 的左子节点`isLeftChild === true`，如节点 5，此时通过：`parent.left = current.right`，删除节点 5
-- 情况 6：current 为父节点 parent 的右子节点`isLeftChild == false`，如节点 9，此时通过：`parent.right = current.right`，删除节点 9
-
-
-
- 
-
-
-##### 4.2.3 要删除的节点，有**两个子节点**
-
-> 这种情况比较**棘手**，让我们先来观察一下，这种情况下有没有什么规律可循
-
-
-
- 
-
-
-**删除节点 9**
-
-在保证删除节点 9 后原二叉树仍为二叉搜索树的前提下，有两种方式：
-
-- 方式 1：从节点 9 的左子树中选择一合适的节点替代节点 9，可知节点 8 符合要求
-- 方式 2：从节点 9 的右子树中选择一合适的节点替代节点 9，可知节点 10 符合要求
-
-
-
- 
-
-
-**删除节点 7**
-
-在保证删除节点 7 后原二叉树仍为二叉搜索树的前提下，也有两种方式：
-
-- 方式 1：从节点 7 的左子树中选择一合适的节点替代节点 7，可知节点 5 符合要求
-- 方式 2：从节点 7 的右子树中选择一合适的节点替代节点 7，可知节点 8 符合要求
-
-
-
- 
-
-
-**删除节点 15**
-
-在保证删除节点 15 后原树二叉树仍为二叉搜索树的前提下，同样有两种方式：
-
-- 方式 1：从节点 15 的左子树中选择一合适的节点替代节点 15，可知节点 14 符合要求
-- 方式 2：从节点 15 的右子树中选择一合适的节点替代节点 15，可知节点 18 符合要求
-
-
-
- 
-
-
-相信你已经发现其中的规律了！
-
-**规律总结：**如果要删除的节点有两个子节点，甚至子节点还有子节点，这种情况下需要从要删除节点**下面的子节点中找到一个合适的节点**，来替换当前的节点。
-
-若用 current 表示需要删除的节点，则合适的节点指的是
-
-- current 左子树中比 current**小一点点的节点**，即 current**左子树**中的**最大值**
-- current 右子树中比 current**大一点点的节点**，即 current**右子树**中的**最小值**
-
-**前驱&后继**
-
-在二叉搜索树中，这两个特殊的节点有特殊的名字：
-
-- 比 current 小一点点的节点，称为 current 节点的**前驱**。比如下图中的节点 5 就是节点 7 的前驱；
-- 比 current 大一点点的节点，称为 current 节点的**后继**。比如下图中的节点 8 就是节点 7 的后继；
-
-
-
- 
-
-
-**代码实现：**
-
-- 查找需要被删除的节点 current 的后继时，需要在 current 的**右子树**中查找**最小值**，即在 current 的**右子树**中一直**向左遍历**查找；
-- 查找前驱时，则需要在 current 的**左子树**中查找**最大值**，即在 current 的**左子树**中一直**向右遍历**查找。
-
-下面只讨论查找 current 后继的情况，查找前驱的原理相同，这里暂不讨论。
-
-```js
-// 获取后继的节点，即从要删除的节点的右边开始查找最小的值
-BinarySearchTree.prototype.getSuccessor = function (delNode) {
- var successor = delNode; // 保存后继
- var current = delNode.right; // 保存删除节点的右子树
- var successorParent = delNode; // 保存后继节点的父节点
- // 循环查找 current 的右子树节点
- while (current !== null) {
- successorParent = successor; // 保存父节点
- successor = current; // 保存后继
- current = current.left;
- }
- // 判断寻找到的后续节点是否直接就是要删除节点的 right
- if (successor !== delNode.right) {
- successorParent.left = successor.right;
- successor.right = delNode.right;
- }
-
- return successor;
-};
-```
-
-###### 代码实现
-
-```js
-// 3.有两个子节点(难点)
-else {
- /**
- * 规律：如果要删除的节点有两个子节点，甚至子节点还有子节点，
- * 这种情况下需要从要删除节点下面的子节点中找到一个合适的节点，来替换当前的节点
- * 若用 current 表示需要删除的节点，则合适的节点指的是：
- * - current 左子树中比 current 小一点点的节点，即 current 左子树中的最大值
- * - current 右子树中比 current 大一点点的节点，即 current 右子树中的最小值
- */
- // 找到后继节点
- var successor = this.getSuccessor(current);
- // 判断是否为根节点
- if (this.root === successor) {
- this.root = successor; // 使用后继替换根节点
- } else if (isLeftChild) {
- parent.left = successor; // 使用后继替换掉 要删除节点的位置
- } else {
- parent.right = successor; // 使用后继替换掉 要删除节点的位置
- }
-
- successor.left = current.left; // 将后继节点的左子树，与要删除节点的左子树相连
-}
-// 删除成功
-return true;
-};
-```
-
-##### 4.2.4 完整的 remove 代码
-
-```js
-BinarySearchTree.prototype.remove = function (key) {
- var current = this.root; // current保存要删除的节点
- var parent = null; // 保存要删除节点的父节点
- var isLeftChild = false; // 用于判断要删除的节点是父节点的左节点还是右节点
-
- // 寻找要删除的节点
- while (current.key != key) {
- parent = current;
- if (key < current.key) {
- isLeftChild = true;
- current = current.left;
- } else {
- isLeftChild = false;
- current = current.right;
- }
- // 若没有找到，说明不需要删除，返回false
- if (current === null) return false;
- }
-
- // 找到了对应的节点,进行情况分析
- // 1.删除的是叶子结点
- if (current.left === null && current.right === null) {
- // 1.1 删除的是根节点
- if (current === this.root) {
- this.root = null;
- } else if (isLeftChild) {
- parent.left = null;
- } else {
- parent.right = null;
- }
- }
-
- // 2.删除的节点只有一个子节点
- else if (current.left === null) {
- if (current === this.root) {
- this.root = current.right;
- } else if (isLeftChild) {
- parent.left = current.right;
- } else {
- parent.right = current.right;
- }
- } else if (current.right === null) {
- if (current === this.root) {
- this.root = current.left;
- } else if (isLeftChild) {
- parent.left = current.left;
- } else {
- parent.right = current.left;
- }
- }
- // 3.有两个子节点(难点)
- else {
- /**
- * 规律：如果要删除的节点有两个子节点，甚至子节点还有子节点，
- * 这种情况下需要从要删除节点下面的子节点中找到一个合适的节点，来替换当前的节点
- * 若用 current 表示需要删除的节点，则合适的节点指的是：
- * - current 左子树中比 current 小一点点的节点，即 current 左子树中的最大值
- * - current 右子树中比 current 大一点点的节点，即 current 右子树中的最小值
- */
- var successor = this.getSuccessor(current);
- if (this.root === successor) {
- this.root = successor;
- } else if (isLeftChild) {
- parent.left = successor;
- } else {
- parent.right = successor;
- }
-
- successor.left = current.left;
- }
-
- return true;
-};
-
-// 获取后继的节点，即从要删除的节点的右边开始查找最小的值
-BinarySearchTree.prototype.getSuccessor = function (delNode) {
- var successor = delNode;
- var current = delNode.right;
- var successorParent = delNode;
-
- while (current !== null) {
- successorParent = successor;
- successor = current;
- current = current.left;
- }
-
- if (successor !== delNode.right) {
- successorParent.left = successor.right;
- successor.right = delNode.right;
- }
-
- return successor;
-};
-```
-
-#### 5. 完整的二叉搜索树代码
-
-```js
-function BinarySearchTree() {
- function Node(key) {
- this.left = null;
- this.key = key;
- this.right = null;
- }
-
- this.root = null;
-
- BinarySearchTree.prototype.insert = function (key) {
- // 1. 创建一个节点类
- var newNode = new Node(key);
- // 2. 判断根节点是否为空
- if (this.root === null) {
- // 2.1 直接插入新的节点
- this.root = newNode;
- } else {
- // 2.2 插入节点类，实现根节点非空时的插入
- this._insertNode(this.root, newNode);
- }
- };
-
- BinarySearchTree.prototype._insertNode = function (node, newNode) {
- // 向左寻找
- if (node.key > newNode.key) {
- if (node.left === null) {
- node.left = newNode;
- } else {
- this._insertNode(node.left, newNode);
- }
- } else {
- // 向右寻找
- if (node.right === null) {
- node.right = newNode;
- } else {
- this._insertNode(node.right, newNode);
- }
- }
- };
-
- //1.先序遍历 ==> 根 -> 左 -> 右
- BinarySearchTree.prototype.preOrderTraversal = function (handler) {
- this._preOrderTraversalNode(this.root, handler);
- };
-
- BinarySearchTree.prototype._preOrderTraversalNode = function (node, handler) {
- if (node !== null) {
- // 1.处理经过的节点
- handler(node.key);
- // 2.处理左节点
- this._preOrderTraversalNode(node.left, handler);
- // 3.处理右节点
- this._preOrderTraversalNode(node.right, handler);
- }
- };
-
- // 2.中序遍历 ==> 左 -> 根 -> 右
- BinarySearchTree.prototype.midOrderTraversal = function (handler) {
- this._midOrderTraversalNode(this.root, handler);
- };
-
- BinarySearchTree.prototype._midOrderTraversalNode = function (node, handler) {
- if (node !== null) {
- // 1.处理左节点
- this._midOrderTraversalNode(node.left, handler);
- // 2.处理经过的节点
- handler(node.key);
- // 3.处理右节点
- this._midOrderTraversalNode(node.right, handler);
- }
- };
-
- // 3.后序遍历 ==> 左 -> 右 -> 根
- BinarySearchTree.prototype.postOrderTraversal = function (handler) {
- this._postOrderTraversalNode(this.root, handler);
- };
-
- BinarySearchTree.prototype._postOrderTraversalNode = function (
- node,
- handler,
- ) {
- if (node !== null) {
- // 1.处理左节点
- this._postOrderTraversalNode(node.left, handler);
- // 2.处理右节点
- this._postOrderTraversalNode(node.right, handler);
- // 3.处理经过的节点
- handler(node.key);
- }
- };
-
- BinarySearchTree.prototype.min = function () {
- var node = this.root;
- while (node.left !== null) {
- node = node.left;
- }
- return node.key;
- };
-
- BinarySearchTree.prototype.max = function () {
- var node = this.root;
- while (node.right !== null) {
- node = node.right;
- }
- return node.key;
- };
-
- BinarySearchTree.prototype.search = function (key) {
- return this._searchNode(this.root, key);
- };
-
- BinarySearchTree.prototype._searchNode = function (node, key) {
- // 这里使用while循环
- var node = this.root;
- while (node !== null) {
- if (key < node.key) {
- node = node.left;
- } else if (key > node.key) {
- node = node.right;
- } else {
- return true;
- }
- }
- return false;
-
- // 递归调用，占用空间，
- // if (node === null) return false;
- // if (node.key < key) {
- // // 查找右边的树
- // this._searchNode(node.right, key);
- // } else if (node.key > key) {
- // //查找左边的树
- // this._searchNode(node.left, key);
- // } else {
- // // 相等情况，说明找到
- // return true;
- // }
- };
-
- BinarySearchTree.prototype.remove = function (key) {
- var current = this.root; // current保存要删除的节点
- var parent = null; // 保存要删除节点的父节点
- var isLeftChild = false; // 用于判断要删除的节点是父节点的左节点还是右节点
-
- // 寻找要删除的节点
- while (current.key != key) {
- parent = current;
- if (key < current.key) {
- isLeftChild = true;
- current = current.left;
- } else {
- isLeftChild = false;
- current = current.right;
- }
- // 若没有找到，说明不需要删除，返回false
- if (current === null) return false;
- }
-
- // 找到了对应的节点,进行情况分析
- // 1.删除的是叶子结点
- if (current.left === null && current.right === null) {
- // 1.1 删除的是根节点
- if (current === this.root) {
- this.root = null;
- } else if (isLeftChild) {
- parent.left = null;
- } else {
- parent.right = null;
- }
- }
-
- // 2.删除的节点只有一个子节点
- else if (current.left === null) {
- if (current === this.root) {
- this.root = current.right;
- } else if (isLeftChild) {
- parent.left = current.right;
- } else {
- parent.right = current.right;
- }
- } else if (current.right === null) {
- if (current === this.root) {
- this.root = current.left;
- } else if (isLeftChild) {
- parent.left = current.left;
- } else {
- parent.right = current.left;
- }
- }
- // 3.有两个子节点(难点)
- else {
- /**
- * 规律：如果要删除的节点有两个子节点，甚至子节点还有子节点，
- * 这种情况下需要从要删除节点下面的子节点中找到一个合适的节点，来替换当前的节点
- * 若用 current 表示需要删除的节点，则合适的节点指的是：
- * - current 左子树中比 current 小一点点的节点，即 current 左子树中的最大值
- * - current 右子树中比 current 大一点点的节点，即 current 右子树中的最小值
- */
- var successor = this.getSuccessor(current);
- if (this.root === successor) {
- this.root = successor;
- } else if (isLeftChild) {
- parent.left = successor;
- } else {
- parent.right = successor;
- }
-
- successor.left = current.left;
- }
-
- return true;
- };
-
- // 获取后继的节点，即从要删除的节点的右边开始查找最小的值
- BinarySearchTree.prototype.getSuccessor = function (delNode) {
- var successor = delNode;
- var current = delNode.right;
- var successorParent = delNode;
-
- while (current !== null) {
- successorParent = successor;
- successor = current;
- current = current.left;
- }
-
- if (successor !== delNode.right) {
- successorParent.left = successor.right;
- successor.right = delNode.right;
- }
-
- return successor;
- };
-}
-
-// 测试代码
-var bst = new BinarySearchTree();
-
-bst.insert(11);
-bst.insert(7);
-bst.insert(15);
-bst.insert(5);
-bst.insert(3);
-bst.insert(9);
-bst.insert(8);
-bst.insert(10);
-bst.insert(13);
-bst.insert(12);
-bst.insert(14);
-bst.insert(20);
-bst.insert(18);
-bst.insert(25);
-bst.insert(6);
-
-// var resString = " ";
-// bst.preOrderTraversal(function (key) {
-// resString += key + " ";
-// });
-// console.log(resString); // ? 11 7 5 3 6 9 8 10 15 13 12 14 20 18 25
-
-// var resString = " ";
-// bst.midOrderTraversal(function (key) {
-// resString += key + " ";
-// });
-// console.log(resString); // ? 3 5 6 7 8 9 10 11 12 13 14 15 18 20 25
-
-// var resString = " ";
-// bst.postOrderTraversal(function (key) {
-// resString += key + " ";
-// });
-// console.log(resString); // ? 3 6 5 8 10 9 7 12 14 13 18 25 20 15 11
-// console.log(bst.max()); // ? 25
-// console.log(bst.min()); // ? 3
-
-// console.log(bst.remove(8));
-// console.log(bst.remove(7));
-// console.log(bst.remove(78));
-
-// var resString = " ";
-// bst.midOrderTraversal(function (key) {
-// resString += key + " ";
-// });
-// console.log(resString); // ?
-```
-
-## 三、平衡树
-
-二叉搜索树的缺陷：当插入的数据是有序的数据，就会造成二叉搜索树的深度过大。
-
-比如原二叉搜索树由 11 7 15 组成，如下图所示：
-
-
-
- 
-
-
-当插入一组有序数据：6 5 4 3 2 就会变成深度过大的搜索二叉树，会严重影响二叉搜索树的性能。
-
-
-
- 
-
-
-### **非平衡树**
-
-- 比较好的二叉搜索树，它的数据应该是**左右均匀分布**的
-- 但是插入**连续数据**后，二叉搜索树中的数据分布就变得**不均匀**了，我们称这种树为**非平衡树**
-- 对于一棵**平衡二叉树**来说，插入/查找等操作的效率是**O（logN）**
-- 而对于一棵**非平衡二叉树**来说，相当于编写了一个链表，查找效率变成了**O（N）**
-
-### **树的平衡性**
-
-为了能以**较快的时间 O（logN）**来操作一棵树，我们需要**保证树总是平衡**的
-
-- 起码大部分是平衡的，此时的时间复杂度也是接近 O（logN）的；
-- 这就要求树中**每个节点左边的子孙节点**的个数，应该尽可能地等于**右边的子孙节点**的个数
-
-### **常见的平衡树**
-
-- **AVL 树**：是最早的一种平衡树，它通过在每个节点多存储一个额外的数据来保持树的平衡。由于 AVL 树是平衡树，所以它的时间复杂度也是 O（logN）。但是它的整体效率不如红黑树，开发中比较少用。
-- **红黑树**：同样通过**一些特性**来保持树的平衡，时间复杂度也是 O（logN）。进行插入/删除等操作时，性能优于 AVL 树，所以平衡树的应用基本都是红黑树。
-
-## 四、关于深度优先和广度优先遍历
-
-实际开发中，我们直接操作二叉树的情况比较少，但是操作下面这种 tree 机构的情况比较多，而我们对于这种树，最常见的遍历方式就是深度优先遍历(DFS)和广度优先遍历(BFS)，所以在这里我简单拓展一下这两种遍历方式
-
-```js
-const tree = {
- val: 'a',
- children: [
- {
- val: 'b',
- children: [
- {
- val: 'd',
- children: [],
- },
- {
- val: 'e',
- children: [],
- },
- ],
- },
- {
- val: 'c',
- children: [
- {
- val: 'f',
- children: [],
- },
- {
- val: 'g',
- children: [],
- },
- ],
- },
- ],
-};
-```
-
-### 1. 深度优先遍历(dfs)
-
-> 这个算法会尽可能深的搜索树的分支。当节点 v 的所在边都己被探寻过，搜索将回溯到发现节点 v 的那条边的起始节点。这一过程一直进行到已发现从源节点可达的所有节点为止。如果还存在未被发现的节点，则选择其中一个作为源节点并重复以上过程，整个进程反复进行直到所有节点都被访问为止
-
-```js
-/**
- * @param {*} root 根结点
- * 思路
- * 访问根节点
- * 对根节点的子节点诶个进行深度优先遍历 ==> 递归
- */
-const dfs = (root) => {
- console.log(root.val);
- root.children.forEach(dfs);
-};
-dfs(tree); // a,b,d,e,c,f,g
-```
-
-### 2. 广度优先遍历(bfs)
-
-> 又译作**宽度优先搜索**，或**横向优先搜索**，是一种[图形搜索算法](https://zh.wikipedia.org/wiki/搜索算法)。简单的说，BFS 是从[根节点](
-
- 
-
-
-### 红黑树的相对平衡
-
-前面 5 条规则的约束确保了以下红黑树的关键特性：
-
-- 从**根到叶子节点**的**最长路径**，不会超过**最短路径**的**两倍**
-- 结果就是这棵树**基本**是平衡的
-- 虽然没有做到绝对的平衡，但是可以保证在最坏的情况下，该树依然是高效的
-
-为什么可以做到**最长路径不超过最短路径的两倍**呢？
-
-- **性质 4**决定了路径上不能有两个相连的红色节点
-- 所以，最长路径一定是红色节点和黑色节点交替而成的
-- 由于根节点和叶子节点都是黑色的，最短路径可能都是黑色节点，并且最长路径中一定是黑色节点多于红色节点
-- **性质 5**决定了所有路径上都有相同数目的黑色节点
-- 这就表明了没有路径能多于其他任何路径两倍长
-
-## 三、红黑树的三种变化
-
-插入一个新节点时，有可能树不再平衡，可以通过三种方式的变换使树保持平衡：
-
-- **变色**
-- **左旋转**
-- **右旋转**
-
-### 3.1 变色
-
-为了重新符合红黑树的规则，需要把**红色**节点变为**黑色**，或者把**黑色**节点变为**红色**
-
-插入的**新节点**通常都是**红色节点**：
-
-- 当插入的节点为**红色**的时候，大多数情况**不违反**红黑树的任何规则
-- 而**插入黑色节点，**必然会导致一条路径上多了一个**黑色节点**，这是很难调整的
-- 红色节点虽然可能导致**红红相连**的情况，但是这种情况可以通过**颜色调换和旋转**来调整
-
-### 3.2 左旋转
-
-以节点 X 为根**逆时针**旋转二叉搜索树，使得父节点原来的位置被自己的右子节点替代，左子节点的位置被父节点替代
-
-
-
- 
-
-
-**旋转过程详解：**
-
-如上图所示，左旋转之后：
-
-- 节点 X 取代了节点 a 原来的位置
-- 节点 Y 取代了节点 X 原来的位置
-- 节点 X 的**左子树** a 仍然是节点 X 的**左子树**（这里 X 的左子树只有一个节点，有多个节点时同样适用，以下同理）
-- 节点 Y 的**右子树** c 仍然是节点 Y 的**右子树**
-- 节点 Y 的**左子树** b 向**左平移**成为了节点 X 的**右子树**
-
-除此之外，二叉搜索树左旋转之后仍为二叉搜索树：
-
-
-
- 
-
-
-### 3.3 右旋转
-
-以节点 X 为根**顺时针**旋转二叉搜索树，使得父节点原来的位置被自己的左子节点替代，右子节点的位置被父节点替代
-
-
-
- 
-
-
-旋转过程详解：
-
-如上图所示，右旋转之后：
-
-- 节点 X 取代了节点 a 原来的位置
-- 节点 Y 取代了节点 X 原来的位置
-- 节点 X 的**右子树** a 仍然是节点 X 的**右子树**（这里 X 的右子树虽然只有一个节点，但是多个节点时同样适用，以下同理）
-- 节点 Y 的**左子树** b 仍然是节点 Y 的**左子树**
-- 节点 Y 的**右子树** c 向**右平移**成为了**节点 X**的**左子树**
-
-除此之外，二叉搜索树右旋转之后仍为二叉搜索树：
-
-
-
- 
-
-
-## 四、红黑树的插入操作
-
-首先需要明确，在保证满足红黑树 5 条规则的情况下，新插入的节点必然是**红色节点**。
-
-为了方便说明，规定以下四个节点：新插入节点为**N**（Node），N 的父节点为**P**（Parent），P 的兄弟节点为**U**（Uncle），U 的父节点为**G**（Grandpa），如下图所示：
-
-
-
- 
-
-
-### 情况 1
-
-当插入的新节点 N 位于树的根上时，没有父节点。
-
-这种情况下，只需要将**红色**节点**变为黑色**节点即可满足规则 2 。
-
-
-
- 
-
-
-### 情况 2
-
-新节点 N 的父节点 P 为黑色节点，此时不需要任何变化。此时既满足规则 4 也满足规则 5。
-
-尽管新节点是红色的，但是新节点 N 有两个黑色节点 NIL，所以通向它的路径上黑色节点的个数依然相等，因此满足规则 5 。
-
-
-
- 
-
-
-### 情况 3
-
-节点 P 为红色，节点 U 也为红色，此时节点 G 必为黑色，即**父红 叔红 祖黑**。在这种情况下需要：
-
-- 先将父节点 P 变为黑色
-- 再将叔叔节点 U 变为黑色
-- 最后将祖父节点 G 变为红色
-
-即变为**父黑叔黑祖红**，如下图所示：
-
-
-
- 
-
-
-可能出现的问题：
-
-- N 的 **祖父节点 G** 的 **父节点** 也可能是 **红色**，这就违反了规则 4，此时可以通过**递归调整节点颜色**
-- 当递归**调整到根节点时**就需要**旋转**了，如下图**节点 A 和节点 B**所示，具体情况后面会介绍
-
-
-
- 
-
-
-### 情况 4
-
-节点 P 是红色节点，节点 U 是黑色节点，并且节点 N 为节点 P 的**左子节点**，此时节点 G 一定是黑色节点，即**父红叔黑祖黑**。
-
-在这种情况下需要：
-
-- **先变色**：将父节点 P 变为黑色，将祖父节点 G 变为红色，即**父黑祖红**
-- **后旋转**：以**祖父节点 G 为根**进行**右旋转**
-
-
-
- 
-
-
-### 情况 5
-
-节点 P 是红色节点，节点 U 是黑色节点，并且节点 N 为节点 P 的**右子节点**，此时节点 G 一定是黑色节点，即**父红叔黑祖黑**。
-
-在这种情况下需要：
-
-- 先以**节点 P 为根进行左旋转，** 旋转后如图 b 所示
-- 随后将**红色节点 P 和黑色节点 B**看成一个整体的**红色节点 N1**，将**新插入的红色节点 N** 看成 **红色节点 P1** 如图 c 所示。此时整体就转换为了情况 4。
-
-
-
- 
-
-
-然后可以按照情况 4 进行处理：
-
-- 先变色：将 N1 节点的父节点 P1 变为黑色，将祖父节点 G 变为红色
-- 后旋转：以祖父节点 G 为根进行右旋转，旋转后如图 e 所示
-- 最后将节点 N1 和 P1 变换回来，完成节点 N 的插入，如图 f 所示
-
-
-
- 
-
-
-## 五、实际插入案例
-
-在二叉树中依次插入节点：10，9，8，7，6，5，4，3，2，1 。
-
-如果直接采用普通的二叉搜索树，节点全部插入后是这样的：
-
-
-
- 
-
-
-是一个严重的**不平衡树**，相当于一个链表，不能体现出二叉搜索树的高效率。而按照红黑树的五条规则插入节点就能最大程度保证搜索二叉树是一棵**平衡树**。
-
-以下为在红黑树中实际插入 10，9，9....1 过程的详解，**为了方便解释省略了部分红黑树的叶子节点（NIL）**
-
-### 插入 10
-
-符合**情况 1**：
-
-- 插入节点 10
-- 将节点 10 的颜色变为黑色
-
-
-
- 
-
-
-### 插入 9
-
-符合**情况 2**：
-
-- 不需要任何变化
-
-
-
- 
-
-
-### 插入 8
-
-我们快速判断属于情况 3 还是情况 4 的方法：
-
-从新插入的**节点 N**出发，按图示箭头经过的四个节点，若为**红红黑红** 三个红色节点则为情况 3，若为**红红黑黑**两个红色节点则为情况 4
-
-
-
- 
-
-
-经过判断，我们插入 8 时符合**情况 4**，所以需要以下操作：
-
-- 父节点 9 变成黑，祖父节点 10 变为红
-- 以祖父节点为根进行右旋转
-
-
-
- 
-
-
-### 插入 7
-
-符合**情况 3**：
-
-- 父节点 8 和叔节点 10 变为黑，祖父节点 9 变为红
-- 此时会出现问题：不符合规则 2，即根节点不为黑，此时可以把以 9 为根节点的二叉搜索树当作一个整体作为一个新插入的节点 N，而此时又符合情况 1，只需要把 9 变回黑色即可
-
-
-
- 
-
-
-### 插入 6
-
-符合**情况 4**：
-
-- 父节点 7 变为黑，祖父节点 8 变为红
-- 以祖父节点 8 为根进行右旋转
-
-
-
- 
-
-
-### 插入 5
-
-符合**情况 3**：
-
-- 父节点 6 和叔节点 8 变为黑，祖父节点 7 变为红
-
-
-
- 
-
-
-### 插入 4
-
-符合**情况 4**：
-
-- 父节点 5 变为黑，祖父节点 6 变为红
-- 以祖父节点 6 为根进行右旋转
-
-
-
- 
-
-
-### 插入 3
-
-**第一次变换**：符合**情况 3**：
-
-- 父节点 4 和叔节点 6 变为黑，祖父节点 5 变为红
-
-变换之后发现 5 和 7 为相连的两个红色节点，于是把以 5 为根的整个子树看成一个新插入的节点 N1，再进行第二次变换。
-
-
-
- 
-
-
-**第二次变换**：符合**情况 4**：
-
-- 父节点 7 变为黑，祖父节点 9 变为红
-- 以祖父节点 9 为根进行右旋转
-
-
-
- 
-
-
-最后复原 N1 得到变换后的红黑树：
-
-
-
- 
-
-
-### 插入 2
-
-符合**情况 4**：
-
-- 父节点 3 变为黑，祖父节点 4 变为红
-- 以祖父节点 4 为根进行右旋转
-
-
-
- 
-
-
-### 插入 1
-
-**第一次变换**：符合**情况 3**：
-
-- 父节点 2 和叔节点 4 变为黑，祖父节点 3 变为红
-
-变换之后发现 3 和 5 为相连的两个红色节点，于是把以 3 为根的整个子树看成一个新插入的节点 N1，再进行第二次变换
-
-
-
- 
-
-
-**第二次变换**：符合**情况 3**：
-
-- **父节点 5 和叔节点 9 变为黑**，**祖父节点 7 变为红**
-
-变换之后发现根节点 7 为红色不符合规则 2，所以把以 7 为根节点的红黑树看成一个新插入的节点 N2，再进行第三次变换
-
-**第三次变换**：符合**情况 1**：
-
-- 直接将根节点 7 变为黑色即可
-
-
-
- 
-
-
-由此，完成了 1~10 节点的插入，虽然没有遇到情况 5，不过情况 5 经过左旋转的操作便可转换为情况 4，原理一样。如下图所示，将这棵红黑树的叶子节点 NIL 补全之后，经检验满足红黑树的五条规则，并且基本属于**平衡树**，效率较高。
-
-
-
- 
-
diff --git a/docs/data-structure/11_graph.md b/docs/data-structure/11_graph.md
deleted file mode 100644
index ffd06a7..0000000
--- a/docs/data-structure/11_graph.md
+++ /dev/null
@@ -1,613 +0,0 @@
----
-nav:
- title: 数据结构
- order: 1
-group:
- title: 常见数据结构
- order: 1
-order: 11
----
-
-# 图(Graph)
-
-## 一、图的介绍
-
-### 1. 什么是图？
-
-- **图结构**是一种与**树结构**有些相似的数据结构
-- **图论**是数学的一个分支，并且，在数学中，树是图的一种
-- 图论以图为研究对象，研究**顶点**和**边**组成的**图形**的数学理论和方法
-- 主要的研究目的为：**事物之间的联系**，**顶点**代表**事物**，**边**代表两个事物间的**关系**
-
-### 2.图的特点
-
-- **一组顶点**：通常用 **V** （Vertex）表示顶点的集合
-- 一组边：通常用 E（Edge）表示边的集合
- - 边是顶点和顶点之间的连线
- - 边可以是有向的，也可以是无向的。比如 A----B 表示无向，A ---> B 表示有向
-
-### 3. 图的常用术语：
-
-- **顶点：**表示图中的一个**节点**
-- **边：**表示**顶点和顶点**给之间的**连线**
-- **相邻顶点：**由一条边连接在一起的顶点称为**相邻顶点**
-- **度：**一个顶点的**度**是**相邻顶点的数量**
-- **路径：**
- - **简单路径：** 简单路径要求不包含重复的顶点
- - **回路：** 第一个顶点和最后一个顶点**相同**的路径称为回路
-- **无向图：** 图中的所有边都是**没有**方向的
-- **有向图：** 图中的所有边都是**有**方向的
-- **无权图：** 无权图中的边没有任何权重意义
-- **带权图：** 带权图中的边有一定的权重含义
-
-### 4. 图的表示
-
-#### 4.1 邻接矩阵
-
-表示图的常用方式为：**邻接矩阵**，如下图所示
-
-- 可以使用二维数组来表示邻接矩阵
-- 邻接矩阵让**每个节点和一个整数相关联**，该**整数作为数组的下标值**
-- 使用一个**二维数组**来表示顶点之间的**连接**
-
-
-
- 
-
-
-邻接矩阵表示法示例，如上图所示：
-
-- 二维数组中的**0**表示**没有连线**，**1**表示**有连线**
- - 如：A[ 0 ] [ 3 ] = 1，表示 A 和 C 之间有连接
-- 邻接矩阵的对角线上的值都为 0，表示 A - A ，B - B，等**自回路都没有连接**（自己与自己之间没有连接）
-- 若为无向图，则邻接矩阵应为对角线上元素全为 0 的对称矩阵
-
-##### 邻接矩阵存在的问题：
-
-- 如果图是一个**稀疏图**，那么邻接矩阵中将存在**大量的 0**，造成**存储空间的浪费**
-
-#### 4.2 邻接表
-
-另外一种表示图的常用方式为：**邻接表**，如下图所示
-
-- 邻接表由图中**每个顶点**以及**和顶点相邻的顶点列表**组成
-- 这个列表可用多种方式存储，比如：**数组/链表/字典（哈希表）** 等都可以
-
-
-
- 
-
-
-如上图所示：
-
-- 图中可清楚看到**A 与 B、C、D 相邻**，假如要表示这些与 A 顶点相邻的顶点（边），可以通过将它们作为 A 的值（value）存入到对应的**数组/链表/字典**中
-- 之后，通过键（key）A，可以十分方便地取出对应的数据
-
-##### 邻接表的问题：
-
-- 邻接表可以简单地得出**出度**，即**某一顶点指向其他顶点的个数**；
-- 但是，邻接表计算**入度**（指向某一顶点的其他顶点的个数称为该顶点的入度）十分困难，此时需要构造**逆邻接表**才能有效计算入度
-
-## 二、图结构的封装
-
-这里我们采用**邻接表**的方式进行封装，使用之前封装过的**字典结构**（也可以理解为 Map）来**存储临近表**
-
-### 2.1 图类的创建
-
-```js
-function Graph() {
- this.vertexes = []; // 存放顶点
- this.adList = new Dictionary(); // 使用字典结构存放 边 信息
-}
-```
-
-### 2.2 添加顶点和边
-
-- 我们需要创建一个数组对象 vertexes 存储图的顶点
-
-- 创建一个字典对象 edges，来存储图的边，其中 key 为顶点值，value 为存储 key 顶点**相邻顶点的数组**
-- 如下图所示
-
-
-
- 
-
-
-```js
-// 1.添加顶点
-Graph.prototype.addVertex = function (val) {
- this.vertexes.push(val); // 添加顶点
- this.adList.set(val, []); // 初始化顶点对应的边
-};
-
-// 2.添加边
-Graph.prototype.addEdge = function (val1, val2) {
- // 因为边是相互存在的，所以需要存储关联两个顶点的信息
- // 这里实现的是无向图，所以不考虑方向的问题
- this.adList.get(val1).push(val2);
- this.adList.get(val2).push(val1);
-};
-```
-
-### 2.3 实现 toString()方法
-
-为图类 Graph 添加 toString 方法，实现以邻接表的形式输出图中各顶点
-
-```js
-// 3.toString方法
-Graph.prototype.toString = function () {
- var resString = '';
- for (var i = 0; i < this.vertexes.length; i++) {
- resString += this.vertexes[i] + '->';
- var adList = this.adList.get(this.vertexes[i]);
- for (var j = 0; j < adList.length; j++) {
- resString += adList[j] + ' ';
- }
- resString += '\n';
- }
- return resString;
-};
-```
-
-#### 测试代码
-
-```js
-// 测试代码
-let graph = new Graph();
-
-// 添加顶点
-let myVertexes = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I'];
-for (let i = 0; i < myVertexes.length; i++) {
- graph.addVertex(myVertexes[i]);
-}
-
-// 添加边
-graph.addEdge('A', 'B');
-graph.addEdge('A', 'C');
-graph.addEdge('A', 'D');
-graph.addEdge('C', 'D');
-graph.addEdge('C', 'G');
-graph.addEdge('D', 'G');
-graph.addEdge('D', 'H');
-graph.addEdge('B', 'E');
-graph.addEdge('B', 'F');
-graph.addEdge('E', 'I');
-
-console.log(graph.toString());
-// A->B C D
-// B->A E F
-// C->A D G
-// D->A C G H
-// E->B I
-// F->B
-// G->C D
-// H->D
-// I->E
-```
-
-### 2.4 图的遍历方式
-
-##### 图的遍历思想：
-
-- 图的遍历思想与树的遍历思想一样，意味着需要将图中**所有的顶点**都访问一遍，并且不能有**重复的访问**（上面的 toString 方法会重复访问）
-
-##### 遍历图的两种算法：
-
-- 广度优先搜索（Breadth - First Search，简称**BFS**）
-- 深度优先搜索（Depth - First Search，简称**DFS**）
-- 两种遍历算法都需要指定**第一个被访问的顶点**
-
-为了记录顶点是否被访问过，这里使用**三种颜色**来表示它们的状态
-
-- **白色**：表示该顶点还没有被访问过
-- **灰色**：表示该顶点被访问过，但其相邻顶点并未完全被访问过
-- **黑色**：表示该顶点被访问过，且其所有相邻顶点都被访问过
-
-首先封装 initializeColor 方法将图中的所有顶点初始化为白色，代码实现如下：
-
-```js
-/**
- * 4.初始化顶点颜色
- * 白色表示该顶点还没有被访问.
- * 灰色表示该顶点被访问过, 但并未被探索过.
- * 黑色表示该顶点被访问过且被完全探索过.
- */
-Graph.prototype._initColors = function () {
- var colors = [];
- for (var i = 0; i < this.vertexes.length; i++) {
- colors[this.vertexes[i]] = 'white';
- }
- return colors;
-};
-```
-
-### 2.5 广度优先搜索(BFS)
-
-#### 思路：
-
-- 广度优先搜索算法会从指定的第一个顶点开始遍历图，先访问其所有的相邻顶点，就像一次访问图的一层
-- 也可以说是**先宽后深**地遍历图中的各个顶点
-
-
-
- 
-
-
-#### **实现思路：**
-
-基于**队列**可以简单地实现广度优先搜索算法：
-
-- 首先创建一个队列 Q（尾部进，首部出）
-- 调用封装的 initializeColor 方法将所有顶点初始化为白色
-- 指定第一个顶点 A，将 A 标注为**灰色**（被访问过的节点），并将 A 放入队列 Q 中
-- 循环遍历队列中的元素，只要队列 Q 非空，就执行以下操作
- - 先将灰色的 A 从 Q 的首部取出
- - 取出 A 后，将 A 的所有未被访问过（白色）的相邻顶点依次从队列 Q 的尾部加入队列，并变为灰色。以此保证，灰色的相邻顶点不重复加入队列
- - A 的全部相邻节点加入 Q 后，A 变为黑色，在下一次循环中被移除 Q 外
-
-#### 代码实现：
-
-```js
-// 5.广度优先遍历
-Graph.prototype.bfs = function (initV, handler) {
- var colors = this._initColors();
- var queue = new Queue();
-
- // 将定点放入队列
- queue.enqueue(initV);
-
- while (!queue.isEmpty()) {
- var qVal = queue.dequeue(); // 拿到队头的顶点
- var qAdj = this.adList.get(qVal); // 拿到队头对应的边
- // 将队头设置正在探索中 灰色
- colors[qVal] = 'gray';
-
- // 将队头顶点相邻的边 全部拿出来遍历一遍，并加入到队列中
- for (var i = 0; i < qAdj.length; i++) {
- var vertexValue = qAdj[i]; // 取出相邻顶点
- // 如果这个顶点是白色，说明没有被访问过
- if (colors[vertexValue] === 'white') {
- // 正在访问
- colors[vertexValue] = 'gray';
- // 将相邻顶点入队
- queue.enqueue(vertexValue);
- }
- }
-
- // 代表该定点已经探索完毕，设置为黑色
- colors[qVal] = 'black';
-
- // 输出
- handler && handler(qVal);
- }
-};
-```
-
-##### 测试代码
-
-```js
-graph.bfs(graph.vertexes[0], function (value) {
- console.log(value); // A,B,C,D,E,F,G,H,I,
-});
-```
-
-#### 过程详解
-
-下为指定的第一个顶点为 A 时的遍历过程：
-
-- 如 a 图所示，将在字典 edges 中取出的与 A 相邻的且未被访问过的白色顶点 B、C、D 放入队列 que 中并变为灰色，随后将 A 变为黑色并移出队列
-- 接着，如图 b 所示，将在字典 edges 中取出的与 B 相邻的且未被访问过的白色顶点 E、F 放入队列 que 中并变为灰色，随后将 B 变为黑色并移出队列
-
-
-
- 
-
-
-- 如 c 图所示，将在字典 edges 中取出的与 C 相邻的且未被访问过的白色顶点 G（A，D 也相邻不过已变为灰色，所以不加入队列）放入队列 que 中并变为灰色，随后将 C 变为黑色并移出队列
-- 接着，如图 d 所示，将在字典 edges 中取出的与 D 相邻的且未被访问过的白色顶点 H 放入队列 que 中并变为灰色，随后将 D 变为黑色并移出队列
-
-
-
- 
-
-
-如此循环直到队列中元素为 0，即所有顶点都变黑并移出队列后才停止，此时图中顶点已被全部遍历
-
-为了更好的理解 BFS，这里提供一份简版 BFS，核心思路都是一样的
-
-#### 简版 BFS
-
-```js
-const graph = {
- 0: [1, 2],
- 1: [2],
- 2: [0, 3],
- 3: [3],
-};
-
-const bfs = (root) => {
- const visited = new Set(); // 用来存储，节点是否访问过
- visited.add(root);
- const q = [root]; //新建队列，根结点入队
-
- while (q.length) {
- let n = q.shift(); // 队列节点出队，拿到队头
- console.log(n); // 访问节点
- // 访问相邻节点，并入队
- graph[n]?.forEach((item) => {
- if (!visited.has(item)) {
- // 若没被访问过
- q.push(item); // 入队
- visited.add(item); // 表示已经访问过
- }
- });
- }
-};
-
-bfs(2); // 2 0 3 1
-```
-
-### 2.6 深度优先搜索(DFS)
-
-#### 思路：
-
-- 深度优先搜索算法将会从指定的第一个顶点开始遍历图，沿着一条路径遍历直到该路径的最后一个顶点都被访问过为止
-- 接着沿原来路径回退并探索下一条路径，即**先深后宽**地遍历图中的各个顶点
-
-
-
- 
-
-
-#### 实现思路：
-
-- 可以使用**栈**结构来实现深度优先搜索算法
-- 深度优先搜索算法的遍历顺序与二叉搜索树中的先序遍历较为相似，同样可以使用**递归**来实现（递归的本质就是**函数栈**的调用）
-
-基于递归实现深度优先搜索算法：定义 dfs 方法用于调用递归方法 dfsVisit，定义 dfsVisit 方法用于递归访问图中的各个顶点。
-
-这里实现 dfs 时，实现一个辅助函数 dfsVisit，方便递归实现 DFS。
-
-在 dfs 方法中：
-
-- 首先，调用 initializeColor 方法将所有顶点初始化为白色
-- 然后，调用 dfsVisit 方法遍历图的顶点
-
-在 dfsVisit 方法中：
-
-- 首先，将传入的指定节点 v 标注为**灰色**
-- 接着，处理顶点 V
-- 然后，访问 V 的相邻顶点
-- 最后，将顶点 v 标注为黑色
-
-#### 代码实现：
-
-```js
-// 6.深度优先遍历
-Graph.prototype.dfs = function (initValue, handler) {
- var colors = this._initColors(); // 初始化顶点颜色
- this.dfsVisit(initValue, colors, handler); // 遍历
-};
-
-Graph.prototype.dfsVisit = function (v, colors, handler) {
- colors[v] = 'gray'; // 访问中
-
- handler && handler(v); // 访问该顶点
-
- var vList = this.adList.get(v); // 获取相邻顶点
- for (var i = 0; i < vList.length; i++) {
- if (colors[vList[i]] === 'white') {
- this.dfsVisit(vList[i], colors, handler); // 递归访问
- }
- }
-
- colors[v] = 'black'; // 访问结束
-};
-```
-
-##### 测试代码
-
-```js
-graph.dfs(graph.vertexes[0], function (value) {
- console.log(value); // A,B,E,I,F,C,D,G,H
-});
-```
-
-#### 过程详解
-
-这里主要解释一下代码中的第 3 步操作：访问指定顶点的相邻顶点。
-
-- 以指定顶点 A 为例，先从储存顶点及其对应相邻顶点的字典对象 edges 中取出由顶点 A 的相邻顶点组成的数组
-
-
-
- 
-
-
-- **第一步**：A 顶点变为灰色，随后进入第一个 for 循环，遍历 A 白色的相邻顶点：B、C、D；在该 for 循环的第 1 次循环中（执行 B），B 顶点满足：colors == "white"，触发递归，重新调用该方法
-- **第二步**：B 顶点变为灰色，随后进入第二个 for 循环，遍历 B 白色的相邻顶点：E、F；在该 for 循环的第 1 次循环中（执行 E），E 顶点满足：colors == "white"，触发递归，重新调用该方法
-- **第三步**：E 顶点变为灰色，随后进入第三个 for 循环，遍历 E 白色的相邻顶点：I；在该 for 循环的第 1 次循环中（执行 I），I 顶点满足：colors == "white"，触发递归，重新调用该方法
-- **第四步**：I 顶点变为灰色，随后进入第四个 for 循环，由于顶点 I 的相邻顶点 E 不满足：colors == "white"，停止递归调用。过程如下图所示：
-
-
-
- 
-
-
-- **第五步**：递归结束后一路向上返回，首先回到第三个 for 循环中继续执行其中的第 2、3...次循环，每次循环的执行过程与上面的同理，直到递归再次结束后，再返回到第二个 for 循环中继续执行其中的第 2、3...次循环....以此类推直到将图的所有顶点访问完为止。
-
-下图为遍历图中各顶点的完整过程：
-
-- **发现**表示访问了该顶点，状态变为**灰色**；
-- **探索**表示既访问了该顶点，也访问了该顶点的全部相邻顶点，状态变为**黑色**；
-- 由于在顶点变为灰色后就调用了处理函数 handler，所以 handler 方法的输出顺序为发现顶点的顺序即：A、B、E、I、F、C、D、G、H 。
-
-
-
- 
-
-
-为了更好的理解 DFS，这里也提供一份简版 DFS
-
-#### 简版 DFS
-
-```js
-const graph = {
- 0: [1, 2],
- 1: [2],
- 2: [0, 3],
- 3: [3],
-};
-
-let set = new Set();
-const dfs = (n) => {
- console.log(n);
- set.add(n);
- graph[n]?.forEach((item) => {
- if (!set.has(item)) {
- dfs(item);
- }
- });
-};
-
-dfs(2); // 2 0 1 3
-```
-
-### 2.7 完整代码
-
-```js
-const Dictionary = require('./dict');
-const Queue = require('../03-队列结构/01-queue');
-
-function Graph() {
- this.vertexes = []; // 存放顶点
- this.adList = new Dictionary(); // 使用字典结构存放 边 信息
-
- // 1.添加顶点
- Graph.prototype.addVertex = function (val) {
- this.vertexes.push(val); // 添加顶点
- this.adList.set(val, []); // 初始化顶点对应的边
- };
-
- // 2.添加边
- Graph.prototype.addEdge = function (val1, val2) {
- // 因为边是相互存在的，所以需要存储关联两个顶点的信息
- // 这里实现的是无向图，所以不考虑方向的问题
- this.adList.get(val1).push(val2);
- this.adList.get(val2).push(val1);
- };
-
- // 3.toString方法
- Graph.prototype.toString = function () {
- var resString = '';
- for (var i = 0; i < this.vertexes.length; i++) {
- resString += this.vertexes[i] + '->';
- var adList = this.adList.get(this.vertexes[i]);
- for (var j = 0; j < adList.length; j++) {
- resString += adList[j] + ' ';
- }
- resString += '\n';
- }
- return resString;
- };
-
- /**
- * 4.初始化顶点颜色
- * 白色表示该顶点还没有被访问.
- * 灰色表示该顶点被访问过, 但并未被探索过.
- * 黑色表示该顶点被访问过且被完全探索过.
- */
- Graph.prototype._initColors = function () {
- var colors = [];
- for (var i = 0; i < this.vertexes.length; i++) {
- colors[this.vertexes[i]] = 'white';
- }
- return colors;
- };
-
- // 5.广度优先遍历
- Graph.prototype.bfs = function (initV, handler) {
- var colors = this._initColors();
- var queue = new Queue();
-
- // 将定点放入队列
- queue.enqueue(initV);
-
- while (!queue.isEmpty()) {
- var qVal = queue.dequeue(); // 拿到队头的顶点
- var qAdj = this.adList.get(qVal); // 拿到队头对应的边
- // 将队头设置正在探索中 灰色
- colors[qVal] = 'gray';
-
- // 将队头顶点相邻的边 全部拿出来遍历一遍，并加入到队列中
- for (var i = 0; i < qAdj.length; i++) {
- var vertexeValue = qAdj[i]; // 取出相邻顶点
- // 如果这个顶点是白色，说明没有被访问过
- if (colors[vertexeValue] === 'white') {
- // 正在访问
- colors[vertexeValue] = 'gray';
- // 将相邻顶点入队
- queue.enqueue(vertexeValue);
- }
- }
-
- // 代表该定点已经探索完毕，设置为黑色
- colors[qVal] = 'black';
-
- // 输出
- handler && handler(qVal);
- }
- };
-
- // 6.深度优先遍历
- Graph.prototype.dfs = function (initValue, handler) {
- var colors = this._initColors(); // 初始化顶点颜色
- this.dfsVisit(initValue, colors, handler); // 遍历
- };
-
- Graph.prototype.dfsVisit = function (v, colors, handler) {
- colors[v] = 'gray'; // 访问中
-
- handler && handler(v); // 访问该顶点
-
- var vList = this.adList.get(v); // 获取相邻顶点
- for (var i = 0; i < vList.length; i++) {
- if (colors[vList[i]] === 'white') {
- this.dfsVisit(vList[i], colors, handler); // 递归访问
- }
- }
-
- colors[v] = 'black'; // 访问结束
- };
-}
-
-// 测试代码
-let graph = new Graph();
-
-// 添加顶点
-let myVertexes = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I'];
-for (let i = 0; i < myVertexes.length; i++) {
- graph.addVertex(myVertexes[i]);
-}
-
-// 添加边
-graph.addEdge('A', 'B');
-graph.addEdge('A', 'C');
-graph.addEdge('A', 'D');
-graph.addEdge('C', 'D');
-graph.addEdge('C', 'G');
-graph.addEdge('D', 'G');
-graph.addEdge('D', 'H');
-graph.addEdge('B', 'E');
-graph.addEdge('B', 'F');
-graph.addEdge('E', 'I');
-
-console.log(graph.toString());
-// A->B C D
-// B->A E F
-// C->A D G
-// D->A C G H
-// E->B I
-// F->B
-// G->C D
-// H->D
-// I->E
-
-graph.bfs(graph.vertexes[0], function (value) {
- console.log(value); // A,B,C,D,E,F,G,H,I,
-});
-
-graph.dfs(graph.vertexes[0], function (value) {
- console.log(value); // A,B,E,I,F,C,D,G,H
-});
-```
diff --git a/docs/data-structure/index.md b/docs/data-structure/index.md
deleted file mode 100644
index 8fc28b9..0000000
--- a/docs/data-structure/index.md
+++ /dev/null
@@ -1,77 +0,0 @@
----
-nav:
- title: 数据结构
- order: 1
-group:
- title: 介绍
- order: -1
----
-
-# 数据结构
-
-本系列文章是作者在 B 站学习完 [《数据结构与算法》](https://www.bilibili.com/video/BV1x7411L7Q7)后，为了进行总结和复习而整理的学习笔记，视频讲解的特别好，给大家安利一波～
-
-## 一、什么是数据结构？
-
-> 数据结构（data structure）是计算机中存储、组织数据的方式。通常情况下，精心选择的数据结构可以带来最优效率的算法。---中文维基百科
-
-一句话总结：数据结构就是在计算机中，**存储和组织数据的方式**。
-
-例如：图书馆中存放了很多书籍，怎样摆放图书既能放很多书，还方便取？
-
-其实我们主要思考两个问题
-
-- 新书怎么插入？
-- 怎么快速对找到指定的书？
-
-1. 如果新书随便摆放，那么在找指定的书时，图书馆那么大，肯定会累死
-2. 按照书名的拼音字母顺序排放
- 1. 新书的插入， 按照字母顺序找到位置，然后插入。
- 2. 书籍的查找：根据字母顺序进行查找
-3. 现实生活中，图书馆会先划分类别，然后在对应类别中，根据字母顺序插入，这样查找效率会更高
-
-### 结论
-
-- 解决问题方法的效率，根据数据的组织方式有关
-- 计算机中存储的数据量相对于图书馆的书籍来说数据量更大，更多
-- 以什么样的方式，来存储和组织我们的数据才能在使用数据时更加方便呢?
-- 这就是数据结构需要考虑的问题。
-
-### 常见的数据结构
-
-- [数组（Array）](https://developer.mozilla.org/zh-CN/docs/Web/JavaScript/Reference/Global_Objects/Array)
-- [栈（Stack）](/data-structure/01_stack)
-- [队列（Queue）](/data-structure/02_queue)
-- [链表（Linked List）](/data-structure/03_linked_list)
-- [哈希表（Hash）](/data-structure/07_hash_table)
-- [树（Tree）](/data-structure/08_tree&binary_tree)
-- [图（Graph）](/data-structure/11_graph)
-- [堆（Heap）](https://zh.wikipedia.org/wiki/%E5%A0%86%E7%A9%8D)
-
-> 注意：数据结构与算法与语言无关，常见的编程语言都有**直接或间接**的使用上述常见的数据结构。
-
-## 二、什么是算法（Algorithm）？
-
-### 算法的定义
-
-- 一个有限指令集，每条指令的描述不依赖于语言
-- 接收一些输入（有些情况下不需要输入）
-- 产生输入
-- 一定在有限步骤之后终止
-
-算法通俗理解：解决问题的办法/步骤逻辑。数据结构的实现，离不开算法。
-
-## 案例
-
-假如杭州和上海之间有一条高架线，高架线长度是 1,000,000 米，有一天高架线中有一米出现了故障，请你想出一种算法，可以快速定位到处问题的地方。
-
-- 线性查找
- - 从上海的起点开始一米一米的排查，最终一定能找到出问题的线段。
- - 但是如果线段在另一头，我们需要排查 1,000,000 次，这是最坏的情况，平均需要 500,000 次。
-- 二分查找
- - 从中间位置开始排查，看一下问题出在杭州到中间位置，还是中间到上海的位置。
- - 查找对应的问题后，再从中间位置分开，重新锁定一般的路程。
- - 最坏的情况，需要多少次可以排查完呢? 最坏的情况是 20 次就可以找到出问题的地方。
- - 怎么计算出来的呢? log(1000000, 2)，以 2 位底，1000000 的对数 ≈ 20。
-
-结论：你会发现，解决问题的办法有很多，但是好的算法对比于差的算法，效率天壤之别。
diff --git a/docs/guide/index.md b/docs/guide/index.md
deleted file mode 100644
index 3d3ccda..0000000
--- a/docs/guide/index.md
+++ /dev/null
@@ -1,49 +0,0 @@
----
-nav:
- title: 指南
- order: 2
----
-
-# 指南
-
-```md
----
-nav: JavaScript题解 #所属nav
-group:
- title: 基础算法 #所属分组 ==>基础算法
- order: 1 #组(基础算法)于其他组的顺序 [...,-1,0,1,...]
-order: 0 # group内顺序
----
-```
-
-:::info{title=题目跳转指引}
-若要新开窗口跳转题目,可以使用`command+click`（macOS）或`ctrl+click`（Linux / Windows）进行跳转
-:::
-
-
-
- 
-
-
-
-
-
-
-
diff --git a/docs/index.md b/docs/index.md
deleted file mode 100644
index b7673d1..0000000
--- a/docs/index.md
+++ /dev/null
@@ -1,23 +0,0 @@
----
-title: 刷题记录
-description: 这是一个leetcode刷题记录📝,目前有300+道题js版本题解,后续会更新其他语言版本,欢迎PR👋
-keywords: [leetcode-js, 前端刷题, leetcode, 刷题记录]
-hero:
- title: 刷题记录
- description: 这是一个leetcode刷题记录📝,目前有300+道题js版本题解,后续会更新其他语言版本,欢迎PR👋
- actions:
- - text: 开始
- link: /js-logs
- - text: GitHub
- link: https://github.com/gaoxiaoduan/leetcode-js
-features:
- - title: 分类
- emoji: 🔖
- description: 题解针对刷题类型进行分类
- - title: 注释
- emoji: 🗒
- description: 大部分题都会增加完善的注释，便于理解&回顾
- - title: n版本解法
- emoji: 🪢
- description: 尝试不同方案的解法，便于总结刷题套路
----
diff --git a/docs/js-logs/BFS.md b/docs/js-logs/BFS.md
deleted file mode 100644
index 7437466..0000000
--- a/docs/js-logs/BFS.md
+++ /dev/null
@@ -1,20 +0,0 @@
----
-nav: JavaScript题解
-group:
- title: 基础算法
- order: 1
-order: 3
-toc: content
----
-
-# BFS
-
-BFS 的核心思想应该不难理解的，就是把一些问题抽象成图，从一个点开始，向四周开始扩散。一般来说，我们写 BFS 算法都是用「队列」这种数据结构，每次将一个节点周围的所有节点加入队列。
-
-BFS 相对 DFS 的最主要的区别是：BFS 找到的路径一定是最短的，但代价就是空间复杂度可能比 DFS 大很多
-
-
-
-
-
-
diff --git a/docs/js-logs/array.md b/docs/js-logs/array.md
deleted file mode 100644
index 198b164..0000000
--- a/docs/js-logs/array.md
+++ /dev/null
@@ -1,86 +0,0 @@
----
-nav: JavaScript题解
-group: 基础数据结构
-order: 0
-toc: content
----
-
-# [数组](https://leetcode.cn/tag/array/problemset/)
-
-数组是在程序设计中，把具有相同类型的若干元素按有序的形式组织起来的一种形式。
-
-作为线性表的实现方式之一，数组中的元素在内存中是**连续**存储的，且每个元素占相同大小的内存。
-
-## 双指针
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-## 前缀和
-
-
-
-
-
-
-
-
-
-## 差分数组
-
-
-
-
-
-## 二位数组的花式遍历技巧
-
-
-
-
-
-
-
-## 岛屿问题
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
diff --git a/docs/js-logs/backtracking.md b/docs/js-logs/backtracking.md
deleted file mode 100644
index 50a57c6..0000000
--- a/docs/js-logs/backtracking.md
+++ /dev/null
@@ -1,52 +0,0 @@
----
-nav: JavaScript题解
-group:
- title: 进阶算法
- order: 2
-order: 1
-toc: content
----
-
-# 回溯算法
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
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-
-
diff --git a/docs/js-logs/binary-search.md b/docs/js-logs/binary-search.md
deleted file mode 100644
index 3437ec7..0000000
--- a/docs/js-logs/binary-search.md
+++ /dev/null
@@ -1,204 +0,0 @@
----
-nav: JavaScript题解
-group:
- title: 基础算法
- order: 1
-order: 0
-toc: content
----
-
-# 二分搜索
-
-## [704.二分查找](https://leetcode.cn/problems/binary-search/)
- 题目跳转指引
-
- 
-
-
-```js
-/**
- * @param {number} n - a positive integer
- * @return {number}
- */
-var hammingWeight = function (n) {
- let res = 0;
- while (n !== 0) {
- n = n & (n - 1);
- res++;
- }
- return res;
-};
-```
-
-## [231. 2 的幂](https://leetcode.cn/problems/power-of-two/)
- 图解
-
- 
-
-
-```js
-/**
- * @param {number[]} nums
- * @return {number}
- */
-var jump = function (nums) {
- let len = nums.length;
- let farthest = 0;
- let end = 0;
- let jump = 0;
- for (let i = 0; i < len - 1; i++) {
- farthest = Math.max(i + nums[i], farthest);
- if (end === i) {
- jump++;
- end = farthest;
- }
- }
- return jump;
-};
-```
-
-动态规划解法
-时间复杂度 O(N^2)
-空间复杂度 O(1)
-
-```js
-// 动态规划
-/**
- * @param {number[]} nums
- * @return {number}
- */
-var jump = function (nums) {
- let len = nums.length;
- let memo = Array.from({ length: len }).fill(len);
- //表示从p处跳到最后位置的最少步数
- const dp = (nums, p) => {
- // base case
- // 当 p 到达最后的位置，不用跳，返回0
- if (p >= len - 1) return 0;
- if (memo[p] !== len) return memo[p];
- let step = nums[p];
- for (let i = 1; i <= step; i++) {
- let subProblum = dp(nums, nums[i] + i);
- memo[p] = Math.min(memo[p], subProblum + 1);
- }
- return memo[p];
- };
-
- return dp(nums, 0);
-};
-```
-
-## [376.摆动序列](https://leetcode.cn/problems/wiggle-subsequence/)
- 图解
-
- 
-
-由于不必按顺序存储，链表在插入数据的时候可以达到 O(1) 的复杂度，但是查找一个节点或者访问特定编号的节点则需要 O(n) 的时间。
-
-## 链表双指针
-
-
-
-
-
-### [21.合并两个有序链表](https://leetcode.cn/problems/merge-two-sorted-lists/)
- Linked List
-
- 
-
-
-```js
-var MyQueue = function () {
- this.stack1 = [];
- this.stack2 = [];
-};
-
-/**
- * @param {number} x
- * @return {void}
- */
-MyQueue.prototype.push = function (x) {
- this.stack1.push(x);
-};
-
-/**
- * @return {number}
- */
-MyQueue.prototype.pop = function () {
- this.peek(); // 先peek，保证stack2不为空
- return this.stack2.pop();
-};
-
-/**
- * @return {number}
- */
-MyQueue.prototype.peek = function () {
- // 当stack2为空时
- if (this.stack2.length === 0) {
- // 将stack1清空
- while (this.stack1.length) {
- const item = this.stack1.pop(); // stack1栈顶
- this.stack2.push(item); // 将stack1栈顶 入栈
- }
- }
- // 返回stack2的栈顶元素
- return this.stack2[this.stack2.length - 1];
-};
-
-/**
- * @return {boolean}
- */
-MyQueue.prototype.empty = function () {
- return this.stack1.length === 0 && this.stack2.length === 0;
-};
-
-/**
- * Your MyQueue object will be instantiated and called as such:
- * var obj = new MyQueue()
- * obj.push(x)
- * var param_2 = obj.pop()
- * var param_3 = obj.peek()
- * var param_4 = obj.empty()
- */
-```
-
-## [678.有效的括号字符串](https://leetcode.cn/problems/valid-parenthesis-string/)
- 图解
-
- 
-
-
-```js
-var MinStack = function () {
- this.stack = [];
- this.minStack = []; // 保存入栈元素的对应的最小值
-};
-
-/**
- * @param {number} val
- * @return {void}
- */
-MinStack.prototype.push = function (val) {
- this.stack.push(val);
- // 维护minStack的栈顶元素最小
- if (
- this.minStack.length === 0 ||
- val <= this.minStack[this.minStack.length - 1]
- ) {
- this.minStack.push(val);
- } else {
- const minStackPeek = this.minStack[this.minStack.length - 1];
- this.minStack.push(minStackPeek);
- }
-};
-
-/**
- * @return {void}
- */
-MinStack.prototype.pop = function () {
- this.stack.pop();
- this.minStack.pop();
-};
-
-/**
- * @return {number}
- */
-MinStack.prototype.top = function () {
- return this.stack[this.stack.length - 1];
-};
-
-/**
- * @return {number}
- */
-MinStack.prototype.getMin = function () {
- return this.minStack[this.minStack.length - 1];
-};
-
-/**
- * Your MinStack object will be instantiated and called as such:
- * var obj = new MinStack()
- * obj.push(val)
- * obj.pop()
- * var param_3 = obj.top()
- * var param_4 = obj.getMin()
- */
-```
-
-
diff --git a/docs/js-logs/string.md b/docs/js-logs/string.md
deleted file mode 100644
index 6e8503a..0000000
--- a/docs/js-logs/string.md
+++ /dev/null
@@ -1,18 +0,0 @@
----
-nav: JavaScript题解
-group: 基础数据结构
-order: 3
-toc: content
----
-
-# 字符串
-
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diff --git a/docs/js-logs/sword-point-offer-II.md b/docs/js-logs/sword-point-offer-II.md
deleted file mode 100644
index 2e070b5..0000000
--- a/docs/js-logs/sword-point-offer-II.md
+++ /dev/null
@@ -1,16 +0,0 @@
----
-nav: JavaScript题解
-group:
- title: TODO
- order: 3
-order: 1
-toc: content
----
-
-# 剑指 Offer II
-
-
-
-
-
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diff --git a/docs/js-logs/sword-point-offer.md b/docs/js-logs/sword-point-offer.md
deleted file mode 100644
index 853850b..0000000
--- a/docs/js-logs/sword-point-offer.md
+++ /dev/null
@@ -1,154 +0,0 @@
----
-nav: JavaScript题解
-group:
- title: TODO
- order: 3
-order: 0
-toc: content
----
-
-# 剑指 Offer
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+ `}))||[],t=(0,v.useState)(0),u=s()(t,2),p=u[0],g=u[1],A=function(o){g(o)};return(0,a.jsxs)("div",{className:"tab-container",children:[(0,a.jsx)("ul",{className:"tab-list",children:x.map(function(c,o){return(0,a.jsx)("li",{className:p===o?"tab-item active":"tab-item",onClick:function(){return A(o)},children:c.props.label},c.props.label)})}),(0,a.jsx)("div",{className:"tab-content",children:(j=x[p])===null||j===void 0||(e=j.props)===null||e===void 0?void 0:e.children})]})},f=l}}]);
diff --git a/dumi__tmp-production__dumi__theme__ContextWrapper.4fbdeb0e.async.js b/dumi__tmp-production__dumi__theme__ContextWrapper.4fbdeb0e.async.js
new file mode 100644
index 0000000..dd15487
--- /dev/null
+++ b/dumi__tmp-production__dumi__theme__ContextWrapper.4fbdeb0e.async.js
@@ -0,0 +1 @@
+"use strict";(self.webpackChunkleetcode_record=self.webpackChunkleetcode_record||[]).push([[923],{77998:function(M,o,e){e.r(o),e.d(o,{default:function(){return v}});var a=e(54306),u=e.n(a),_=e(93236),t=e(5899),c=e(13036),n=e(96481),m=e(90237),i=e(62086),E={};function v(){var D=(0,t.pC)(),f=(0,_.useState)(!0),s=u()(f,2),p=s[0],d=s[1],r=(0,_.useRef)(t.m8.location.pathname);return(0,_.useEffect)(function(){return t.m8.listen(function(l){l.location.pathname!==r.current&&(r.current=l.location.pathname,d(!0),document.documentElement.scrollTo(0,0))})},[]),(0,i.jsx)(c.D.Provider,{value:{pkg:{name:"leetcode-record",description:"This is a leetcode brush record",version:"0.0.1",license:"MIT",authors:["1961018000@qq.com"]},entryExports:E,demos:n.DE,components:n.wx,locales:m.k,loading:p,setLoading:d,themeConfig:{footer:" \u{1F4AB} keep awake \u{1F4AB}",prefersColor:{default:"light",switch:!0},name:"leetcode-js",logo:"/leetcode-js/logo.png"}},children:D})}}}]);
diff --git "a/example/BFS/752.\346\211\223\345\274\200\350\275\254\347\233\230\351\224\201.md" "b/example/BFS/752.\346\211\223\345\274\200\350\275\254\347\233\230\351\224\201.md"
deleted file mode 100644
index 7509cc2..0000000
--- "a/example/BFS/752.\346\211\223\345\274\200\350\275\254\347\233\230\351\224\201.md"
+++ /dev/null
@@ -1,79 +0,0 @@
-## [752.打开转盘锁](https://leetcode.cn/problems/open-the-lock/)
- 图解115.最小栈
-
-
-
-
-```js
-var CQueue = function () {
- this.stack1 = []; // |==
- this.stack2 = []; // |==
-};
-
-/**
- * @param {number} value
- * @return {void}
- */
-CQueue.prototype.appendTail = function (value) {
- this.stack1.push(value);
-};
-
-/**
- * @return {number}
- */
-CQueue.prototype.deleteHead = function () {
- if (this.stack2.length === 0) {
- while (this.stack1.length > 0) {
- this.stack2.push(this.stack1.pop());
- }
- }
-
- const res = this.stack2.pop();
- return res === undefined ? -1 : res;
-};
-
-/**
- * Your CQueue object will be instantiated and called as such:
- * var obj = new CQueue()
- * obj.appendTail(value)
- * var param_2 = obj.deleteHead()
- */
-```
diff --git "a/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer10-I.\346\226\220\346\263\242\351\202\243\345\245\221\346\225\260\345\210\227.md" "b/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer10-I.\346\226\220\346\263\242\351\202\243\345\245\221\346\225\260\345\210\227.md"
deleted file mode 100644
index 346aba2..0000000
--- "a/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer10-I.\346\226\220\346\263\242\351\202\243\345\245\221\346\225\260\345\210\227.md"
+++ /dev/null
@@ -1,28 +0,0 @@
-## [剑指 Offer 10- I. 斐波那契数列](https://leetcode.cn/problems/fei-bo-na-qi-shu-lie-lcof/)
- 图解
-
-
-
-
-```js
-/**
- * @param {number} n - a positive integer
- * @return {number}
- */
-var hammingWeight = function (n) {
- let res = 0;
- while (n !== 0) {
- n = n & (n - 1);
- res++;
- }
- return res;
-};
-```
diff --git "a/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer16.\346\225\260\345\200\274\347\232\204\346\225\264\346\225\260\346\254\241\346\226\271.md" "b/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer16.\346\225\260\345\200\274\347\232\204\346\225\264\346\225\260\346\254\241\346\226\271.md"
deleted file mode 100644
index 49bfb7d..0000000
--- "a/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer16.\346\225\260\345\200\274\347\232\204\346\225\264\346\225\260\346\254\241\346\226\271.md"
+++ /dev/null
@@ -1,28 +0,0 @@
-## [剑指 Offer 16. 数值的整数次方](https://leetcode.cn/problems/shu-zhi-de-zheng-shu-ci-fang-lcof/)
- 图解
-
- 
-
-
-- 思路
-- 此题只允许原地修改，且结果有序，所以优先想到中序遍历的方法
-- 要购构造双向链表，就要对两个节点进行互相引用
-- 所以遍历过程中需要使用`pre`保存前一个节点，然后更改left和right，保持pre和当前节点互相引用
-- 结果需要双向链表的head和tail也相互引用，所以，遍历过程中要保存head节点
-- tail节点在遍历结束时，pre就是tail节点
-- 最后更改头尾节点的left和right，互相引用，返回head即可
-- 跟[426题](https://leetcode-cn.com/problems/convert-binary-search-tree-to-sorted-doubly-linked-list/)相似
-
-```js
-/**
- * @param {Node} root
- * @return {Node}
- */
-var treeToDoublyList = function (root) {
- if (root === null) return null;
- let head = null, pre = null;
- const dfs = (root) => {
- if (root === null) return;
- // --前序位置--
- dfs(root.left);
- // --中序位置--
- if (pre === null) {
- // 如果pre在中序遍历为null，说明，走到了最左边
- // 也就是走到了头
- head = root;
- } else {
- // 回退时
- // pre为1节点
- // root为2节点
- // 1.right 指向2
- // 2.left 指向1
- // 构建双向链表
- pre.right = root;
- root.left = pre;
- }
- pre = root;
- // --后序位置--
- dfs(root.right);
- }
-
- dfs(root);
- // 头尾互指，构成一个更大的双向脸部
- head.left = pre;
- pre.right = head;
-
- return head;
-};
-```
diff --git "a/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer37.\345\272\217\345\210\227\345\214\226\344\272\214\345\217\211\346\240\221.md" "b/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer37.\345\272\217\345\210\227\345\214\226\344\272\214\345\217\211\346\240\221.md"
deleted file mode 100644
index d8211fd..0000000
--- "a/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer37.\345\272\217\345\210\227\345\214\226\344\272\214\345\217\211\346\240\221.md"
+++ /dev/null
@@ -1,61 +0,0 @@
-## [剑指 Offer 37. 序列化二叉树](https://leetcode.cn/problems/xu-lie-hua-er-cha-shu-lcof/)
- 图解
-
- 
-
-
-跟[233. 数字 1 的个数](/js-logs/math#233-数字-1-的个数)相同
-
-```js
-/**
- * @param {number} n
- * @return {number}
- */
-var countDigitOne = function (n) {
- let digit = 1; // 位因子
- let res = 0;
-
- for (let k = 0; digit <= n; k++) {
- let forwardDigit = digit * 10;
- res +=
- Math.floor(n / forwardDigit) * digit +
- Math.min(Math.max((n % forwardDigit) - digit + 1, 0), digit);
- digit *= 10;
- }
- return res;
-};
-```
diff --git "a/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer44.\346\225\260\345\255\227\345\272\217\345\210\227\344\270\255\346\237\220\344\270\200\344\275\215\347\232\204\346\225\260\345\255\227.md" "b/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer44.\346\225\260\345\255\227\345\272\217\345\210\227\344\270\255\346\237\220\344\270\200\344\275\215\347\232\204\346\225\260\345\255\227.md"
deleted file mode 100644
index 70b396c..0000000
--- "a/example/\345\211\221\346\214\207Offer/\345\211\221\346\214\207Offer44.\346\225\260\345\255\227\345\272\217\345\210\227\344\270\255\346\237\220\344\270\200\344\275\215\347\232\204\346\225\260\345\255\227.md"
+++ /dev/null
@@ -1,37 +0,0 @@
-## [剑指 Offer 44. 数字序列中某一位的数字](https://leetcode.cn/problems/shu-zi-xu-lie-zhong-mou-yi-wei-de-shu-zi-lcof/)
- 公式
-
- 
-
-
-```js
-/**
- * @param {number[][]} matrix
- * @return {number}
- */
-var countSquares = function (matrix) {
- let count = 0;
- const m = matrix.length,
- n = matrix[0].length;
- // 我们用 f[i][j] 表示以(i, j) 为右下角的正方形的最大边长，
- // 除此定义之外，f[i][j] = x 也表示以(i, j) 为右下角的正方形的数目为 x（即边长为 1, 2, ..., x 的正方形各一个）。
- // 在计算出所有的 f[i][j] 后，我们将它们进行累加，就可以得到矩阵中正方形的数目
- const dp = new Array(m).fill(0).map((_) => new Array(n).fill(0));
-
- for (let i = 0; i < m; i++) {
- for (let j = 0; j < n; j++) {
- if (i === 0 || j === 0) {
- dp[i][j] = matrix[i][j];
- } else if (matrix[i][j] === 0) {
- dp[i][j] = 0;
- } else {
- dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1;
- }
- count += dp[i][j];
- }
- }
- return count;
-};
-```
diff --git "a/example/\345\212\250\346\200\201\350\247\204\345\210\222/139.\345\215\225\350\257\215\346\213\206\345\210\206.md" "b/example/\345\212\250\346\200\201\350\247\204\345\210\222/139.\345\215\225\350\257\215\346\213\206\345\210\206.md"
deleted file mode 100644
index 3ddcf2b..0000000
--- "a/example/\345\212\250\346\200\201\350\247\204\345\210\222/139.\345\215\225\350\257\215\346\213\206\345\210\206.md"
+++ /dev/null
@@ -1,41 +0,0 @@
-## [139.单词拆分](https://leetcode.cn/problems/word-break/)
- dp数组图
-
- 
-
-
-```js
-/**
- * @param {string} s
- * @return {string[]}
- */
-var letterCasePermutation = function (s) {
- const res = [];
- const isNumber = (c) => parseInt(c).toString() === 'NaN' ? false : true;
-
- const backTrack = (start, track, sArr) => {
- track.push(sArr.join(''));
- for (let i = start; i < sArr.length; i++) {
- const c = sArr[i];
- if (isNumber(c)) continue;
- // 'a' ==> 97
- // 'A' ==> 65
- if (c >= 'a') {
- // 小写
- sArr[i] = c.toUpperCase();
- backTrack(i + 1, track, sArr);
- sArr[i] = c.toLowerCase();
- } else {
- // 大写
- sArr[i] = c.toLowerCase();
- backTrack(i + 1, track, sArr);
- sArr[i] = c.toUpperCase();
- }
- }
- }
- backTrack(0, res, [...s]);
- return res;
-};
-```
-
-广度优先搜索
-- 思路
-- 时间复杂度 O(n * 2^n)
-- 空间复杂度 O(n * 2^n)
-- 使用一个队列存储搜索的元素
-- 累加队列的第一个元素，当元素的长度等于s的长度，就是一个组合
-- 当s[p]是一个字母，就把这个字母转换一下大小写，然后当做一个新组合入队
-- 当s[p]是一个数字，直接累加到队头元素即可
-
-```js
-var letterCasePermutation = function (s) {
- const ans = [];
- const q = [""];
- while (q.length !== 0) {
- // console.log(q);
- const cur = q[0];
- const p = cur.length;
- if (p === s.length) {
- ans.push(cur);
- q.shift();
- } else {
- if (isLetter(s[p])) {
- q.push(cur + swapCase(s[p]));
- }
- q[0] += s[p];
- }
- }
- return ans;
-};
-
-const swapCase = (ch) => {
- if ('a' <= ch && ch <= 'z') {
- return ch.toUpperCase();
- }
- if ('A' <= ch && ch <= 'Z') {
- return ch.toLowerCase();
- }
-}
-
-const isLetter = (ch) => {
- return 'a' <= ch && ch <= 'z' || 'A' <= ch && ch <= 'Z';
-}
-```
diff --git "a/example/\345\233\236\346\272\257/79.\345\215\225\350\257\215\346\220\234\347\264\242.md" "b/example/\345\233\236\346\272\257/79.\345\215\225\350\257\215\346\220\234\347\264\242.md"
deleted file mode 100644
index fa39878..0000000
--- "a/example/\345\233\236\346\272\257/79.\345\215\225\350\257\215\346\220\234\347\264\242.md"
+++ /dev/null
@@ -1,56 +0,0 @@
-## [79.单词搜索](https://leetcode.cn/problems/word-search/)
- 图解
-
- 
-
-
-```js
-/**
- * @param {string} s
- * @return {string}
- */
-var reverseWords = function (s) {
- let trimStr = [];
- // 去除前后空格和 中间的相邻空格
- for (const c of s.trim()) {
- if (c !== ' ') {
- trimStr.push(c)
- } else if (trimStr.length !== 0 && trimStr[trimStr.length - 1] !== ' ') {
- trimStr.push(' ')
- }
- }
-
- // 翻转arr [i,j]的内容
- const reverse = (arr, i, j) => {
- while (i < j) {
- const temp = arr[i];
- arr[i] = arr[j];
- arr[j] = temp;
- i++;
- j--;
- }
- }
- // 先反转整体
- const n = trimStr.length;
- reverse(trimStr, 0, n - 1);
- // 再翻转每个单词
-
- let start = 0, end = 0;
- while (start < n) {
- while (end < n && trimStr[end] != ' ') {
- end++;
- }
- // while结束，end指向空格
-
- // 翻转单个单词
- reverse(trimStr, start, end - 1);
- start = end + 1; // 将start定位到新的开头[end+1]的位置
- end++; // end也指向新的开头
- }
- // join
- return trimStr.join('')
-};
-```
-
diff --git "a/example/\345\255\227\347\254\246\344\270\262/387.\345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\347\254\254\344\270\200\344\270\252\345\224\257\344\270\200\345\255\227\347\254\246.md" "b/example/\345\255\227\347\254\246\344\270\262/387.\345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\347\254\254\344\270\200\344\270\252\345\224\257\344\270\200\345\255\227\347\254\246.md"
deleted file mode 100644
index 5368629..0000000
--- "a/example/\345\255\227\347\254\246\344\270\262/387.\345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\347\254\254\344\270\200\344\270\252\345\224\257\344\270\200\345\255\227\347\254\246.md"
+++ /dev/null
@@ -1,31 +0,0 @@
-## [387.字符串中的第一个唯一字符](https://leetcode.cn/problems/first-unique-character-in-a-string/)
- 图解
-
-
-
-
-跟[剑指 Offer43.1 ～ n 整数中 1 出现的次数](/js-logs/sword-point-offer#剑指-offer-43-1n-整数中-1-出现的次数)相同
-
-```js
-/**
- * @param {number} n
- * @return {number}
- */
-var countDigitOne = function (n) {
- let digit = 1; // 位因子
- let res = 0;
-
- for (let k = 0; digit <= n; k++) {
- let forwardDigit = digit * 10;
- res +=
- Math.floor(n / forwardDigit) * digit +
- Math.min(Math.max((n % forwardDigit) - digit + 1, 0), digit);
- digit *= 10;
- }
- return res;
-};
-```
diff --git "a/example/\346\225\260\345\255\246/400.\347\254\254N\344\275\215\346\225\260\345\255\227.md" "b/example/\346\225\260\345\255\246/400.\347\254\254N\344\275\215\346\225\260\345\255\227.md"
deleted file mode 100644
index 7c4124a..0000000
--- "a/example/\346\225\260\345\255\246/400.\347\254\254N\344\275\215\346\225\260\345\255\227.md"
+++ /dev/null
@@ -1,35 +0,0 @@
-## [400.第 N 位数字](https://leetcode.cn/problems/nth-digit/)
- 公式
-
- 
-
-
-```js
-/**
- * @param {number[][]} mat
- * @param {number} k
- * @return {number[][]}
- */
-var matrixBlockSum = function (mat, k) {
- const m = mat.length,
- n = mat[0].length;
- let res = new Array(m).fill().map((_) => new Array(n).fill(0));
-
- const numMatrix = new NumMatrix(mat);
-
- for (let i = 0; i < m; i++) {
- for (let j = 0; j < n; j++) {
- // 左上角坐标（x1,y1）
- let x1 = Math.max(i - k, 0);
- let y1 = Math.max(j - k, 0);
- // 右下角的坐标
- let x2 = Math.min(i + k, m - 1);
- let y2 = Math.min(j + k, n - 1);
- res[i][j] = numMatrix.sumRegion(x1, y1, x2, y2);
- }
- }
-
- return res;
-};
-// 304题目
-/**
- * @param {number[][]} matrix
- */
-var NumMatrix = function (matrix) {
- let m = matrix.length,
- n = matrix[0].length;
- // 定义preSum[i,j]记录着 matrix[0][0] 到matrix[i-1][j-1]的元素和(前缀和)
- this.preSum = new Array(m + 1).fill(0).map((_) => new Array(n + 1).fill(0));
- for (let i = 1; i < m + 1; i++) {
- for (let j = 1; j < n + 1; j++) {
- this.preSum[i][j] =
- this.preSum[i - 1][j] +
- this.preSum[i][j - 1] +
- matrix[i - 1][j - 1] -
- this.preSum[i - 1][j - 1];
- }
- }
-};
-
-/**
- * @param {number} row1
- * @param {number} col1
- * @param {number} row2
- * @param {number} col2
- * @return {number}
- */
-NumMatrix.prototype.sumRegion = function (row1, col1, row2, col2) {
- return (
- this.preSum[row2 + 1][col2 + 1] -
- this.preSum[row1][col2 + 1] -
- this.preSum[row2 + 1][col1] +
- this.preSum[row1][col1]
- );
-};
-
-/**
- * Your NumMatrix object will be instantiated and called as such:
- * var obj = new NumMatrix(matrix)
- * var param_1 = obj.sumRegion(row1,col1,row2,col2)
- */
-```
diff --git "a/example/\346\225\260\347\273\204/\345\211\215\347\274\200\345\222\214/238.\351\231\244\350\207\252\350\272\253\344\273\245\345\244\226\346\225\260\347\273\204\347\232\204\344\271\230\347\247\257.md" "b/example/\346\225\260\347\273\204/\345\211\215\347\274\200\345\222\214/238.\351\231\244\350\207\252\350\272\253\344\273\245\345\244\226\346\225\260\347\273\204\347\232\204\344\271\230\347\247\257.md"
deleted file mode 100644
index a1c80f9..0000000
--- "a/example/\346\225\260\347\273\204/\345\211\215\347\274\200\345\222\214/238.\351\231\244\350\207\252\350\272\253\344\273\245\345\244\226\346\225\260\347\273\204\347\232\204\344\271\230\347\247\257.md"
+++ /dev/null
@@ -1,48 +0,0 @@
-### [238.除自身以外数组的乘积](https://leetcode.cn/problems/product-of-array-except-self/)
- 矩阵区域和
-
- 
-
-
-时间复杂度 O(1)
-空间复杂度 O(N)
-
-- 思路
-- 前缀和 解法
-- 初始化是计算前缀和 preSum,preSum 第一位存储 0，方便后面计算，所以长度要比 nums+1
-- nums: 0 1 2 3 4 5
-- ---> : [ 3 5 2 -2 4 1 ]
-- preSum:0 1 2 3 4 5 6
-- -----> : [ 0 3 8 10 8 12 13 ]
-- 核心- 思路是我们 new 一个新的数组 preSum 出来，preSum[i] 记录 nums[0..i-1] 的累加和,看图 10 = 3 + 5 + 2：
-- 如果我想求索引区间 [1, 4] 内的所有元素之和，就可以通过 preSum[5] - preSum[1] 得出
-
-```js
-/**
- * @param {number[]} nums
- */
-var NumArray = function (nums) {
- this.preSum = new Array(nums.length + 1).fill(0);
- for (let i = 1; i < this.preSum.length; i++) {
- this.preSum[i] = this.preSum[i - 1] + nums[i - 1];
- }
-};
-
-/**
- * @param {number} left
- * @param {number} right
- * @return {number}
- */
-NumArray.prototype.sumRange = function (left, right) {
- return this.preSum[right + 1] - this.preSum[left];
-};
-```
-
-- 根据题意求和
-
-```js
-/**
- * @param {number[]} nums
- */
-var NumArray = function (nums) {
- this.nums = nums;
-};
-
-/**
- * @param {number} left
- * @param {number} right
- * @return {number}
- */
-NumArray.prototype.sumRange = function (left, right) {
- let res = 0;
- for (let i = left; i <= right; i++) {
- res += this.nums[i];
- }
- return res;
-};
-```
diff --git "a/example/\346\225\260\347\273\204/\345\211\215\347\274\200\345\222\214/304.\344\272\214\347\273\264\345\214\272\345\237\237\345\222\214\346\243\200\347\264\242-\347\237\251\351\230\265\344\270\215\345\217\257\345\217\230.md" "b/example/\346\225\260\347\273\204/\345\211\215\347\274\200\345\222\214/304.\344\272\214\347\273\264\345\214\272\345\237\237\345\222\214\346\243\200\347\264\242-\347\237\251\351\230\265\344\270\215\345\217\257\345\217\230.md"
deleted file mode 100644
index f43cb91..0000000
--- "a/example/\346\225\260\347\273\204/\345\211\215\347\274\200\345\222\214/304.\344\272\214\347\273\264\345\214\272\345\237\237\345\222\214\346\243\200\347\264\242-\347\237\251\351\230\265\344\270\215\345\217\257\345\217\230.md"
+++ /dev/null
@@ -1,58 +0,0 @@
-### [304.二维区域和检索-矩阵不可变](https://leetcode.cn/problems/range-sum-query-2d-immutable/)
- 前缀和
-
- 
-
-
-- 定义 preSum[i,j]记录着 matrix[0][0] 到 matrix[i-1][j-1]的元素和(前缀和)
-
-```js
-/**
- * @param {number[][]} matrix
- */
-var NumMatrix = function (matrix) {
- let m = matrix.length,
- n = matrix[0].length;
- // 定义preSum[i,j]记录着 matrix[0][0] 到matrix[i-1][j-1]的元素和(前缀和)
- this.preSum = new Array(m + 1).fill(0).map((_) => new Array(n + 1).fill(0));
- for (let i = 1; i < m + 1; i++) {
- for (let j = 1; j < n + 1; j++) {
- this.preSum[i][j] =
- this.preSum[i - 1][j] +
- this.preSum[i][j - 1] +
- matrix[i - 1][j - 1] -
- this.preSum[i - 1][j - 1];
- }
- }
-};
-
-/**
- * @param {number} row1
- * @param {number} col1
- * @param {number} row2
- * @param {number} col2
- * @return {number}
- */
-NumMatrix.prototype.sumRegion = function (row1, col1, row2, col2) {
- return (
- this.preSum[row2 + 1][col2 + 1] -
- this.preSum[row1][col2 + 1] -
- this.preSum[row2 + 1][col1] +
- this.preSum[row1][col1]
- );
-};
-
-/**
- * Your NumMatrix object will be instantiated and called as such:
- * var obj = new NumMatrix(matrix)
- * var param_1 = obj.sumRegion(row1,col1,row2,col2)
- */
-```
diff --git "a/example/\346\225\260\347\273\204/\345\217\214\346\214\207\351\222\210/167.\344\270\244\346\225\260\344\271\213\345\222\214-ii-\350\276\223\345\205\245\346\234\211\345\272\217\346\225\260\347\273\204.md" "b/example/\346\225\260\347\273\204/\345\217\214\346\214\207\351\222\210/167.\344\270\244\346\225\260\344\271\213\345\222\214-ii-\350\276\223\345\205\245\346\234\211\345\272\217\346\225\260\347\273\204.md"
deleted file mode 100644
index 278e3bc..0000000
--- "a/example/\346\225\260\347\273\204/\345\217\214\346\214\207\351\222\210/167.\344\270\244\346\225\260\344\271\213\345\222\214-ii-\350\276\223\345\205\245\346\234\211\345\272\217\346\225\260\347\273\204.md"
+++ /dev/null
@@ -1,37 +0,0 @@
-### [167.两数之和-ii-输入有序数组](https://leetcode.cn/problems/two-sum-ii-input-array-is-sorted/)
- 前缀和的矩阵运算
-
- 
-
-
-其中有四个岛屿，但是左下角和右上角的岛屿形状相同，所以不同的岛屿共有三个，算法返回 3。
-
-- 很显然我们得想办法把二维矩阵中的「岛屿」进行转化，变成比如字符串这样的类型
-- 然后利用 Set 这样的数据结构去重，最终得到不同的岛屿的个数。
-- 1, 2, 3, 4 代表上下左右，用 -1, -2, -3, -4 代表上下左右的撤销
-
-```js
-var numDistinctIslands = function (grid) {
- const m = grid.length,
- n = grid[0].length;
-
- // 淹没与 (i, j) 相邻的陆地，并返回淹没的陆地面积
- const dfs = (i, j, path, direction) => {
- if (i < 0 || i >= m || j < 0 || j >= n) return 0; // 越界
- if (grid[i][j] === 0) return 0; // 已经是海水了
-
- grid[i][j] = 0; // 用海水覆盖
-
- // 前序遍历位置
- path.push(String(direction) + ',');
- // 覆盖四周
- dfs(i - 1, j, path, 1); // 上
- dfs(i + 1, j, path, 2); // 下
- dfs(i, j - 1, path, 3); // 左
- dfs(i, j + 1, path, 4); // 右
- // 后续位置
- path.push(String(-direction) + ',');
- };
-
- let set = new Set();
- for (let i = 0; i < m; i++) {
- for (let j = 0; j < n; j++) {
- // 如果发现岛屿
- if (grid[i][j] === 1) {
- let path = [];
- dfs(i, j, path, 666); // 开始位置不影响正确性
- set.add(path.join(''));
- }
- }
- }
- return set.size;
-};
-
-// 测试代码
-const grid = [
- [1, 1, 0, 1, 1],
- [1, 0, 0, 0, 0],
- [0, 0, 0, 0, 0],
- [0, 0, 0, 0, 1],
- [1, 1, 0, 1, 1],
-];
-const res = numDistinctIslands(grid);
-
-console.log(res, res === 3 ? true : false);
-```
diff --git "a/example/\346\225\260\347\273\204/\345\262\233\345\261\277\351\227\256\351\242\230/695.\345\262\233\345\261\277\347\232\204\346\234\200\345\244\247\351\235\242\347\247\257.md" "b/example/\346\225\260\347\273\204/\345\262\233\345\261\277\351\227\256\351\242\230/695.\345\262\233\345\261\277\347\232\204\346\234\200\345\244\247\351\235\242\347\247\257.md"
deleted file mode 100644
index c15b7b1..0000000
--- "a/example/\346\225\260\347\273\204/\345\262\233\345\261\277\351\227\256\351\242\230/695.\345\262\233\345\261\277\347\232\204\346\234\200\345\244\247\351\235\242\347\247\257.md"
+++ /dev/null
@@ -1,36 +0,0 @@
-### [695.岛屿的最大面积](https://leetcode.cn/problems/max-area-of-island/)
-