leetcode Longest Valid Parentheses

Longest Valid Parentheses

 Total Accepted: 4153 Total Submissions: 22587My Submissions

Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.

For "(()", the longest valid parentheses substring is "()", which has length = 2.

Another example is ")()())", where the longest valid parentheses substring is "()()", which has length = 4.

每次操作栈和其他容器的时候都要注意是否为空。

这道题的计算要注意，如何才能得到当前最长valid parentheses。

int longestValidParentheses(string s) {stack<int> stk;int len = 0;for (int i = 0; i < s.length(); i++){if (s[i] == '(') stk.push(i);else{if (stk.empty()) stk.push(i);else{if (s[stk.top()] == '('){stk.pop();if (stk.empty()) len = max(len, i+1);else len = max(len, i-stk.top());}else stk.push(i);}}}return len;}

巧妙地利用栈，保留下标，然后根据条件判断，逐步计算出结果。

分支条件有三到四个，也挺容易出错的题目。

[cpp] view plaincopyprint?

1. class Solution {  
2. public:  
3.     int longestValidParentheses(string s)   
4.     {  
5.         int n = s.length();  
6.         stack<int> stk;  
7.         int longestLen = 0;  
8.         for (int i = 0; i < n; i++)  
9.         {  
10.             if (s[i] == '(') stk.push(i);  
11.             else  
12.             {  
13.                 if (stk.empty())  
14.                 {  
15.                     stk.push(i);  
16.                 }  
17.                 else  
18.                 {  
19.                     if (s[stk.top()] != '(')  
20.                         stk.push(i);  
21.                     else  
22.                     {  
23.                         stk.pop();  
24.                         //居然会漏了max写成maxLen = (maxLen, i+1)  
25.                         if (stk.empty()) longestLen = max(longestLen, i+1);  
26.                         else    longestLen = max(longestLen, i-stk.top());  
27.                     }  
28.                 }  
29.             }  
30.         }  
31.         return longestLen;  
32.     }  
33. };