[Java] LeetCode32 Longest Valid Parentheses

标签：leetcode

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.

题意：给定一串包含括号的字符串，求字长有效的括号串。能顺序匹配的。这一题刚开始想好久~~，用stack来做题，很容易求出字符串中所有有效的括号长度。但是如何求有效子串呢？如果我们能找出无效括号的index，将有效的index减去上一个无效的index，那么就是该字符的有效子串。弄清楚这一点求最大的有效字符串也会变得很容易。

 public int longestValidParentheses(String s) {
        if(s==null)return 0;
        int len=s.length();
        int i=0;
        Stack<Integer> stack=new Stack<Integer>();
        char ch;
        int res=0;
        while(i<len)
        {
            ch=s.charAt(i);
            if(ch=='(')
            stack.push(i);//我们变换思路，将括号的index入栈
            else
            {
               if(!stack.isEmpty()&&s.charAt(stack.peek())=='(')//如果是’)'，且与stack顶括号匹配时，弹出
               {
                   stack.pop();
                   res=Math.max(stack.isEmpty()?i+1:i-stack.peek(),res);//为空的话，证明前面没有无效括号，将i+1;不为空的，前面有无效字符，减去无效字符的index
               }else
               {
                   stack.push(i);
               }
            }
            i++;
        }
       return res; 
    }

[Java] LeetCode32 Longest Valid Parentheses

标签：leetcode

原文地址：http://blog.csdn.net/fumier/article/details/45331861