leetcode 32. Longest Valid Parentheses DP

第一种方法是dp方法，第二种是利用栈的方法

这道题利用栈的思路很强，虽然普遍都会想到利用栈，但是利用每次进栈的元素来更新空差值，下标之差很精妙，反而dp方法显得不是那么巧妙

万事跪在一个巧字，用下标入栈，秒掉渣渣。

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

Example 1:

Input: "(()"
Output: 2
Explanation: The longest valid parentheses substring is "()"

Example 2:

Input: ")()())"
Output: 4
Explanation: The longest valid parentheses substring is "()()"

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.

   int longestValidParentheses(string s) {
            if(s.length() <= 1) return 0;
            int curMax = 0;
            vector<int> longest(s.size(),0);
            for(int i=1; i < s.length(); i++){
                if(s[i] == ')'){
                    if(s[i-1] == '('){
                        longest[i] = (i-2) >= 0 ? (longest[i-2] + 2) : 2;
                        curMax = max(longest[i],curMax);
                    }
                    else{ // if s[i-1] == ')', combine the previous length.
                        if(i-longest[i-1]-1 >= 0 && s[i-longest[i-1]-1] == '('){
                            longest[i] = longest[i-1] + 2 + ((i-longest[i-1]-2 >= 0)?longest[i-longest[i-1]-2]:0);
                            curMax = max(longest[i],curMax);
                        }
                    }
                }
                //else if s[i] == '(', skip it, because longest[i] must be 0
            }
            return curMax;
        }

解释：想找到最大值，无非就是（（））和（）（）类似于这两种，而a[i]只有是‘）’这种才有讨论情况，否则以‘（’为结尾的不存在，也就是0

当a[i]==')',看以a[i-2]为结尾的有多长，longest[i]=longest[i-2] + 2

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s[i-longest[i-1]-1] == '('

当s[i]是‘）’时，并且s[i-1]是‘）；那么久看以s[i-1]为结尾的有多少，然后减去，i-longest[i-1]-1，看着个是否是‘（’，如果是的话，那就看

i-longest[i-1]-2这个位置的longest，加上就好，前提是i-longest[i-1]-2>0





2.

class Solution {
public:
    int longestValidParentheses(string s) {
        stack<int> stk;
        stk.push(-1);
        int maxL=0;
        for(int i=0;i<s.size();i++)
        {
            int t=stk.top();
            if(t!=-1&&s[i]==')'&&s[t]=='(')
            {
                stk.pop();
                maxL=max(maxL,i-stk.top());
            }
            else
                stk.push(i);
        }
        return maxL;
    }
};

说明：这个栈存放的是原字符串的下标，t栈顶指针，当栈顶存放的下表t=stack.top(),string[t]为‘（’且a[i]为‘）’，就出栈然后更新max 值，他这个原理是什么呢 就是a[i]和栈顶之间的元素全部出战，i-stack.pop() 比如（（）（（（）））这样的

第一个（入展，然后第二个，紧接着发生配对，pop出来，栈顶元素为小标0，而发生配对现象下表已经到8

2-0=max 6-4=max； 7-3

leetcode 32. Longest Valid Parentheses DP

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