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.

思路分析：这题做括号匹配，很容易想到用栈，但是这题栈里面保存括号的index比较好，方便长度计算。基本做法是，遇到左括号入栈，遇到右括号出栈，当进行出栈操作的时候，考虑两种情况

1 当前栈为空，那么没法匹配这个右括号，以该位置的下一个位置i+1为考察括号子串的起点。

2 当前栈不为空，弹出栈顶元素。弹出后如果栈为空，那么合法子串长度是i-start+1;弹出后栈不为空，也就是还有多余的左括号在里面，那么合法子串长度为从当前栈顶元素下一个元素开始，到i结束，也就是i-(stack.peek()+1)+1 = i - stack.peek()。

对字符串进行一遍扫描，时间复杂度和空间复杂度都是O（n）.

AC Code

public class Solution {
    public int longestValidParentheses(String s) {
        //0914
        if(s == null || s.length() < 2) return 0;
        char[] sCharArray = s.toCharArray();
        int n = sCharArray.length;
        Stack stack = new Stack();
        int start = 0;
        int max = 0;
        for(int i = 0; i < n; i++){
            if(sCharArray[i] == '('){
                stack.push(i);
            }else{
                if(stack.isEmpty()){
                    start = i+1;
                } else{
                    stack.pop();
                    max = stack.isEmpty()?Math.max(max, i - start +1):Math.max(max, i - stack.peek());     
                }
            }
        }
        return max;
        //0920
    }
}