*LeetCode-Longest Valid Parentheses

用一个stack记录每个char的位置 假如有可以凑成一对的就pop 没有就在里面留着 这样扫过一遍之后 留下的都是没有凑对的 然后找到所有没有凑对的之间的距离 就是之间连续的valid（）的长度 注意最后一个stack里面的char和string最后的位置 以及stack里面第一个char和0位置之间的两个substring也要计入

public class Solution {
public int longestValidParentheses(String s) {
if ( s == null || s.length() == 0 )
return 0;
Stack <Integer> stack = new Stack <Integer> ();
int length = 0;
for ( int i = 0; i < s.length(); i ++ ){
if ( s.charAt ( i ) == '(' )
stack.push( i );
else {
if ( !stack.isEmpty() && s.charAt(stack.peek()) == '(' ){
stack.pop();
}
else
stack.push ( i );
}
}
int right = s.length();
int left = 0;
while ( !stack.isEmpty () ){
left = stack.pop();
length = Math.max ( length, right - left - 1);
right = left;
}
length = Math.max( length, right );
return length;
}
}

还有一种是dp的做法

My solution uses DP. The main idea is as follows: I construct a array
longest[], for any longest[i], it stores the longest length of valid parentheses which is end at i. 
And the DP idea is : 
If s[i] is '(', set longest[i] to 0,because any string end with '(' cannot be a valid one. 
Else if s[i] is ')' 
     If s[i-1] is '(', longest[i] = longest[i-2] + 2 
     Else if s[i-1] is ')' and s[i-longest[i-1]-1] == '(', longest[i] = longest[i-1] + 2 + longest[i-longest[i-1]-2] 
For example, input "()(())", at i = 5, longest array is [0,2,0,0,2,0], longest[5] = longest[4] + 2 + longest[1] = 6.