4.1.2 Longest Valid Parentheses

原题链接： https://oj.leetcode.com/problems/longest-valid-parentheses/

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.

我的思路是：和前一题Valid Parentheses一样，代码如下：

public class Solution {    public int longestValidParentheses(String s) {        Stack<Character> stack = new Stack<Character>();        int length = 0;        int maxLength = 0;         char c;        for(int i = 0; i < s.length(); i++){            c = s.charAt(i);            if(c == '(' ){                stack.push(c);            }            if(c == ')'){                if(stack.isEmpty()){                   continue;                }                char d = stack.peek();                if(d == '(' && c!= ')') {                    length = 0;                    continue;                }                else {                    stack.pop();                    length = length + 2;                    if(length > maxLength){                        maxLength = length;                    }                }            }        }        return maxLength;    }}

但是这个是不对的。

Input:"()(()"Output:4Expected:2Input:"()(())"Output:4Expected:6

为什么？

正确的代码：

方法I： 栈。Time O(n), Space O(n)

思路： “所有无法匹配的')'”的index其实都是各个group的分界点” 

（引用：http://www.cnblogs.com/lichen782/p/leetcode_Longest_Valid_Parentheses.html）

public class Solution {    public int longestValidParentheses(String s) {        //code after reading dai's c++ code        //each ) is used to separate groups of valid parentheses        Stack<Integer> lefts = new Stack<Integer>();//store the indices of  '('        int length = 0;        int maxLen = 0;         int last = -1;        char c;        for(int i = 0; i < s.length(); i++){            c = s.charAt(i);            if(c == '('){                lefts.push(i);            }            else{//c == ')'                if(lefts.isEmpty()){//this ) should be used as the separation point                    last = i;                }                else{                    lefts.pop();//match the current ) with previous ()                    if(lefts.isEmpty()){//this group has completed                        maxLen = Math.max(maxLen, i - last);                    }                    else{                        maxLen = Math.max(maxLen, i - lefts.peek());                    }                }            }        }        return maxLen;    }}

方法II: DP 不懂。todo

方法III: 两边扫描。不懂为什么work. todo.