Merge branch 'master' of https://github.com/azl397985856/leetcode

luzhipeng · luzhipeng · commit 2b74951fff99 · 2019-06-06T10:01:29.000+08:00
diff --git a/README.en.md b/README.en.md
@@ -184,12 +184,13 @@ The data structures mainly includes:
 
 #### Hard
 - [0023.merge-k-sorted-lists](./problems/23.merge-k-sorted-lists.md)
+- [0032.longest-valid-parentheses](./problems/32.longest-valid-parentheses.md) 🆕
 - [0042.trapping-rain-water](./problems/42.trapping-rain-water.md)
-- [0128.longest-consecutive-sequence](./problems/128.longest-consecutive-sequence.md) 🆕 
+- [0128.longest-consecutive-sequence](./problems/128.longest-consecutive-sequence.md) 🆕
 - [0145.binary-tree-postorder-traversal](./problems/145.binary-tree-postorder-traversal.md)
 - [0146.lru-cache](./problems/146.lru-cache.md)
 - [0239.sliding-window-maximum](./problems/239.sliding-window-maximum.md)
-- [0295.find-median-from-data-stream](./problems/295.find-median-from-data-stream.md) 🆕 
+- [0295.find-median-from-data-stream](./problems/295.find-median-from-data-stream.md) 🆕
 - [0301.remove-invalid-parentheses](./problems/301.remove-invalid-parentheses.md)
 
 
diff --git a/README.md b/README.md
@@ -182,12 +182,13 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 
 #### 困难难度
 - [0023.merge-k-sorted-lists](./problems/23.merge-k-sorted-lists.md)
+- [0032.longest-valid-parentheses](./problems/32.longest-valid-parentheses.md) 🆕
 - [0042.trapping-rain-water](./problems/42.trapping-rain-water.md)
-- [0128.longest-consecutive-sequence](./problems/128.longest-consecutive-sequence.md) 🆕 
+- [0128.longest-consecutive-sequence](./problems/128.longest-consecutive-sequence.md) 🆕
 - [0145.binary-tree-postorder-traversal](./problems/145.binary-tree-postorder-traversal.md)
 - [0146.lru-cache](./problems/146.lru-cache.md)
 - [0239.sliding-window-maximum](./problems/239.sliding-window-maximum.md)
-- [0295.find-median-from-data-stream](./problems/295.find-median-from-data-stream.md) 🆕 
+- [0295.find-median-from-data-stream](./problems/295.find-median-from-data-stream.md) 🆕
 - [0301.remove-invalid-parentheses](./problems/301.remove-invalid-parentheses.md)
 
 ### 数据结构与算法的总结
@@ -253,4 +254,4 @@ http://t.me/leetcode_intl
 
 - 如果有想法和创意，请提[issue](https://github.com/azl397985856/leetcode/issues)或者进群提
 - 如果想贡献代码，请提[PR](https://github.com/azl397985856/leetcode/pulls)
-- 如果需要修改项目中图片，[这里](./assets/drawio/)存放了项目中绘制图的源代码， 大家可以用[draw.io](https://www.draw.io/)打开进行编辑。
+- 如果需要修改项目中图片，[这里](./assets/drawio/)存放了项目中绘制图的源代码， 大家可以用[draw.io](https://www.draw.io/)打开进行编辑。
diff --git a/assets/drawio/32.longest-valid-parentheses.drawio b/assets/drawio/32.longest-valid-parentheses.drawio
@@ -0,0 +1 @@
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diff --git a/assets/problems/32.longest-valid-parentheses.png b/assets/problems/32.longest-valid-parentheses.png
diff --git a/problems/102.binary-tree-level-order-traversal.md b/problems/102.binary-tree-level-order-traversal.md
@@ -30,6 +30,10 @@ return its level order traversal as:
 
 然后不断的出队， 如果出队的是null，则表式这一层已经结束了，我们就继续push一个null。
 
+如果不入队特殊元素Null来表示每层的结束，则在while循环开始时保存当前队列的长度，以保证每次只遍历一层（参考下面的C++ Code）。
+
+> 如果采用递归方式，则需要将当前节点，当前所在的level以及结果数组传递给递归函数。在递归函数中，取出节点的值，添加到level参数对应结果数组元素中（参考下面的C++ Code）。
+
 
 ## 关键点解析
 
@@ -43,7 +47,9 @@ return its level order traversal as:
 
 
 ## 代码
+* 语言支持：JS，C++
 
+Javascript Code:
 ```js
 /*
  * @lc app=leetcode id=102 lang=javascript
@@ -123,7 +129,66 @@ var levelOrder = function(root) {
 
   return items;
 };
+
+```
+C++ Code:
+```C++
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+ 
+ // 迭代
+class Solution {
+public:
+    vector<vector<int>> levelOrder(TreeNode* root) {
+        auto ret = vector<vector<int>>();
+        if (root == nullptr) return ret;
+        auto q = vector<TreeNode*>();
+        q.push_back(root);
+        auto level = 0;
+        while (!q.empty())
+        {
+            auto sz = q.size();
+            ret.push_back(vector<int>());
+            for (auto i = 0; i < sz; ++i)
+            {
+                auto t = q.front();
+                q.erase(q.begin());
+                ret[level].push_back(t->val);
+                if (t->left != nullptr) q.push_back(t->left);
+                if (t->right != nullptr) q.push_back(t->right);
+            }
+            ++level;
+        }
+        return ret;
+    }
+};
+
+// 递归
+class Solution {
+public:
+    vector<vector<int>> levelOrder(TreeNode* root) {
+        vector<vector<int>> v;
+        levelOrder(root, 0, v);
+        return v;
+    }
+private:
+    void levelOrder(TreeNode* root, int level, vector<vector<int>>& v) {
+        if (root == NULL) return;
+        if (v.size() < level + 1) v.resize(level + 1);
+        v[level].push_back(root->val);
+        levelOrder(root->left, level + 1, v);
+        levelOrder(root->right, level + 1, v);
+    }
+};
 ```
+
 ## 相关题目
 - [103.binary-tree-zigzag-level-order-traversal](./103.binary-tree-zigzag-level-order-traversal.md)
 - [104.maximum-depth-of-binary-tree](./104.maximum-depth-of-binary-tree.md)
diff --git a/problems/104.maximum-depth-of-binary-tree.md b/problems/104.maximum-depth-of-binary-tree.md
@@ -44,12 +44,14 @@ var maxDepth = function(root) {
 
 - 队列
 
-- 队列中用 Null(一个特殊元素)来划分每层
+- 队列中用 Null(一个特殊元素)来划分每层，或者在对每层进行迭代之前保存当前队列元素的个数（即当前层所含元素个数）
 
 - 树的基本操作- 遍历 - 层次遍历（BFS）
 
 ## 代码
+* 语言支持：JS，C++
 
+JavaScript Code:
 ```js
 /*
  * @lc app=leetcode id=104 lang=javascript
@@ -94,6 +96,40 @@ var maxDepth = function(root) {
   return depth;
 };
 ```
+C++ Code:
+```C++
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    int maxDepth(TreeNode* root) {
+        if (root == nullptr) return 0;
+        auto q = vector<TreeNode*>();
+        auto d = 0;
+        q.push_back(root);
+        while (!q.empty())
+        {
+            ++d;
+            auto sz = q.size();
+            for (auto i = 0; i < sz; ++i)
+            {
+                auto t = q.front();
+                q.erase(q.begin());
+                if (t->left != nullptr) q.push_back(t->left);
+                if (t->right != nullptr) q.push_back(t->right);
+            }
+        }
+        return d;
+    }
+};
+```
 ## 相关题目
 - [102.binary-tree-level-order-traversal](./102.binary-tree-level-order-traversal.md)
-- [103.binary-tree-zigzag-level-order-traversal](./103.binary-tree-zigzag-level-order-traversal.md)
+- [103.binary-tree-zigzag-level-order-traversal](./103.binary-tree-zigzag-level-order-traversal.md)
diff --git a/problems/11.container-with-most-water.md b/problems/11.container-with-most-water.md
@@ -66,6 +66,9 @@ eg:
 
 
 ## 代码
+* 语言支持：JS，C++
+
+JavaScript Code:
 
 ```js
 /*
@@ -135,4 +138,19 @@ var maxArea = function(height) {
     return max;
 };
 ```
-
+C++ Code:
+```C++
+class Solution {
+public:
+    int maxArea(vector<int>& height) {
+        auto ret = 0ul, leftPos = 0ul, rightPos = height.size() - 1;
+        while( leftPos < rightPos)
+        {
+            ret = std::max(ret, std::min(height[leftPos], height[rightPos]) * (rightPos - leftPos));
+            if (height[leftPos] < height[rightPos]) ++leftPos;
+            else --rightPos;
+        }
+        return ret;
+    }
+};
+```
diff --git a/problems/32.longest-valid-parentheses.md b/problems/32.longest-valid-parentheses.md
@@ -0,0 +1,81 @@
+## 题目地址
+https://leetcode.com/problems/longest-valid-parentheses/
+
+## 题目描述
+
+```
+Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.
+
+Example 1:
+
+Input: "(()"
+Output: 2
+Explanation: The longest valid parentheses substring is "()"
+Example 2:
+
+Input: ")()())"
+Output: 4
+Explanation: The longest valid parentheses substring is "()()"
+```
+
+## 思路(动态规划)
+
+所有的动态规划问题, 首先需要解决的就是如何寻找合适的子问题.
+该题需要我们找到最长的有效括号对, 我们首先想到的就是定义**dp[i]为前i个字符串的最长有效括号对长度**, 但是随后我们会发现, 这样的定义, 我们无法找到dp[i]和dp[i-1]的任何关系.
+所以, 我们需要重新找一个新的定义: 定义**dp[i]为以第i个字符结尾的最长有效括号对长度**. 然后, 我们通过下面这个例子找一下dp[i]和dp[i-1]之间的关系.
+
+```python
+s = '(())())'
+```
+
+从上面的例子我们可以观察出一下几点结论(**描述中i为图中的dp数组的下标, 对应s的下标应为i-1, 第i个字符的i从1开始**).
+1. base case: 空字符串的最长有效括号对长度肯定为0, 即: dp[0] = 0;
+2. s的第**1**个字符结尾的最长有效括号对长度为0, s的第**2**个字符结尾的最长有效括号对长度也为0, 这个时候我们可以得出结论: 最长有效括号对不可能以'('结尾, 即: dp[1] = d[2] = 0;
+3. 当i等于3时, 我们可以看出dp[2]=0, dp[3]=2, 因为第2个字符(**s[1]**)和第3个字符(**s[2]**)是配对的;
+   当i等于4时, 我们可以看出dp[3]=2, dp[4]=4, 因为我们配对的是第1个字符(**s[0]**)和第4个字符(**s[3]**);
+   因此, 我们可以得出结论: 如果第**i**个字符和第<strong>i-1-dp[i-1]</strong>个字符是配对的, 则dp[i] = dp[i-1] + 2, 其中: i-1-dp[i-1] >= 1, 因为第0个字符没有任何意义;
+4. 根据第3条规则来计算的话, 我们发现dp[5]=0, dp[6]=2, 但是显然, dp[6]应该为6才对, 但是我们发现可以将"(())"和"()"进行拼接, 即: dp[i] += dp[i-dp[i]], 即: dp[6] = 2 + dp[6-2] = 2 + dp[4] = 6
+
+根据以上规则, 我们求解dp数组的结果为: [0, 0, 0, 2, 4, 0, 6, 0], 其中最长有效括号对的长度为6. 以下为图解:
+![32.longest-valid-parentheses](../assets/problems/32.longest-valid-parentheses.png)
+
+## 关键点解析
+
+1. 第3点特征, 需要检查的字符是s[i-1]和s[i-2-dp[i-1]], 根据定义可知: i-1 >= dp[i-1], 但是i-2不一定大于dp[i-1], 因此, 需要检查越界;
+3. 第4点特征最容易遗漏, 还有就是不需要检查越界, 因为根据定义可知: i >= dp[i], 所以dp[i-dp[i]]的边界情况是dp[0];
+
+## 代码
+
+* 语言支持: Python
+
+Python Code:
+```
+class Solution:
+    def longestValidParentheses(self, s: str) -> int:
+        mlen = 0
+        slen = len(s)
+        dp = [0] * (slen + 1)
+        for i in range(1, len(s) + 1):
+            # 有效的括号对不可能会以'('结尾的
+            if s[i - 1] == '(':
+                continue
+
+            left_paren = i - 2 - dp[i - 1]
+            if left_paren >= 0 and s[left_paren] == '(':
+                dp[i] = dp[i - 1] + 2
+
+                # 拼接有效括号对
+                if dp[i - dp[i]]:
+                    dp[i] += dp[i - dp[i]]
+
+                # 更新最大有效扩对长度
+                if dp[i] > mlen:
+                    mlen = dp[i]
+
+        return mlen
+```
+
+## 扩展
+
+1. 如果判断的不仅仅只有'('和')', 还有'[', ']', '{'和'}', 该怎么办?
+2. 如果输出的不是长度, 而是任意一个最长有效括号对的字符串, 该怎么办?
diff --git a/thinkings/basic-data-structure-en.md b/thinkings/basic-data-structure-en.md
diff --git a/thinkings/basic-data-structure.md b/thinkings/basic-data-structure.md

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