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Merge pull request doocs#257 from matteokjh/feature/lcof_js
feat: add javaScript solutions for lcof problems
2 parents 82a3f00 + 429392a commit ebda970

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‎lcof/面试题03. 数组中重复的数字/README.md

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@@ -59,6 +59,21 @@ class Solution {
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}
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```
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### JavaScript
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```js
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/**
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* @param {number[]} nums
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* @return {number}
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*/
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var findRepeatNumber = function(nums) {
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let m = {}
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for(let num of nums) {
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if(m[num]) return num
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m[num] = 1
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}
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};
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```
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### ...
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```
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‎lcof/面试题03. 数组中重复的数字/Solution.js

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/**
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* @param {number[]} nums
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* @return {number}
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*/
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var findRepeatNumber = function(nums) {
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let m = {}
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for(let num of nums) {
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if(m[num]) return num
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m[num] = 1
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}
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};

‎lcof/面试题04. 二维数组中的查找/README.md

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}
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```
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### JavaScript
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```js
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/**
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* @param {number[][]} matrix
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* @param {number} target
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* @return {boolean}
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*/
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var findNumberIn2DArray = function(matrix, target) {
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let row = matrix.length
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let col = matrix[0].length
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function dfs(i,j) {
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if(i < 0 || j >= col) {
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return false
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}
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if(matrix[i][j] === target) return true
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else if(matrix[i][j] > target) {
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return dfs(i-1,j)
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} else {
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return dfs(i,j+1)
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}
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}
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return dfs(row-1,0)
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};
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```
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### ...
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```
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‎lcof/面试题04. 二维数组中的查找/Solution.js

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/**
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* @param {number[][]} matrix
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* @param {number} target
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* @return {boolean}
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*/
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var findNumberIn2DArray = function(matrix, target) {
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let row = matrix.length
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let col = matrix[0].length
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function dfs(i,j) {
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if(i < 0 || j >= col) {
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return false
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}
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if(matrix[i][j] === target) return true
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else if(matrix[i][j] > target) {
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return dfs(i-1,j)
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} else {
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return dfs(i,j+1)
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}
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}
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return dfs(row-1,0)
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};

‎lcof/面试题05. 替换空格/README.md

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}
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```
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### JavaScript
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```js
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/**
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* @param {string} s
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* @return {string}
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*/
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var replaceSpace = function(s) {
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return s.split(' ').join('%20')
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};
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```
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### ...
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```
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‎lcof/面试题05. 替换空格/Solution.js

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/**
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* @param {string} s
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* @return {string}
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*/
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var replaceSpace = function(s) {
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return s.split(' ').join('%20')
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};

‎lcof/面试题06. 从尾到头打印链表/README.md

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}
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```
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### JavaScript
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```js
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```
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### ...
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```
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‎lcof/面试题06. 从尾到头打印链表/Solution.js

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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* @param {ListNode} head
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* @return {number[]}
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*/
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var reversePrint = function(head) {
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let node = head
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let res = []
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while(node) {
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res.unshift(node.val)
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node = node.next
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}
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return res
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};

‎lcof/面试题07. 重建二叉树/README.md

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}
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```
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### JavaScript
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```js
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/**
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* Definition for a binary tree node.
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* function TreeNode(val) {
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* this.val = val;
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* this.left = this.right = null;
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* }
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*/
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/**
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* @param {number[]} preorder
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* @param {number[]} inorder
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* @return {TreeNode}
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*/
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var buildTree = function(preorder, inorder) {
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if(!preorder || !preorder.length) return null
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let preIdx = 0
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let inMap = {}
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for(let i=0;i<inorder.length;i++) {
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inMap[inorder[i]] = i
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}
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function func(start, end) {
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if(start > end) {
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return null
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}
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let preVal = preorder[preIdx]
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preIdx++
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let inIdx = inMap[preVal]
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let node = new TreeNode(preVal)
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node.left = func(start, inIdx - 1)
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node.right = func(inIdx + 1, end)
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return node
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}
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return func(0, preorder.length - 1)
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};
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```
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### ...
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```
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‎lcof/面试题07. 重建二叉树/Solution.js

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/**
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* Definition for a binary tree node.
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* function TreeNode(val) {
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* this.val = val;
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* this.left = this.right = null;
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* }
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*/
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/**
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* @param {number[]} preorder
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* @param {number[]} inorder
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* @return {TreeNode}
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*/
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var buildTree = function(preorder, inorder) {
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if(!preorder || !preorder.length) return null
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let preIdx = 0
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let inMap = {}
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for(let i=0;i<inorder.length;i++) {
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inMap[inorder[i]] = i
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}
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function func(start, end) {
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if(start > end) {
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return null
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}
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let preVal = preorder[preIdx]
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preIdx++
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let inIdx = inMap[preVal]
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let node = new TreeNode(preVal)
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node.left = func(start, inIdx - 1)
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node.right = func(inIdx + 1, end)
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return node
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}
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return func(0, preorder.length - 1)
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};

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