【leetcode】 Longest Valid Parentheses (hard)★

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：自己按照早上Scramble String (hard)★ 的动态规划方法试着做了一遍，‘（’表示为1， ‘）’表示为-1，用dp[start]记录不同长度情况下以start开始的括号对应的数字之和。如果为0则表示有效，更新最大长度。 结果超时了，说明我这个O(n²)的方法不行。

    int longestValidParentheses(string s) {
        string scpy = s;
        int ans = 0;
        while(!scpy.empty() && scpy[0] == ')') //跳过位于前面的 ) 因为肯定无法配对
        {
            scpy.erase(scpy.begin());
        }
        if(scpy.empty())
            return ans;

        vector<int> dp;  //dp[start]
        vector<int> dp1;
        for(int start = 0; start <= scpy.length() - 1; start++)
        {
            dp.push_back((scpy[start] == '(') ? 1 : -1);
            dp1.push_back(dp.back());
        }
        for(int len = 2; len <= scpy.length(); len++)
        {
            for(int start = 0; start <=scpy.length() - len; start++)
            {
                dp[start] = dp[start] + dp1[start + len - 1];
                if(dp[start] == 0 && len > ans)
                {
                    ans = len;
                }
            }
        }
        return ans;
    }

大神可以通过而且简洁的O(n)方法

用longest[i]存储以 i 结尾时最大有效长度（必须包括第 i 个字符）

如果s[i] = '('   longest[i] = 0

else s[i] = ')'  

          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] 

 the condition "s[i-1] == '('" since "s[i-longest[i-1]-1] == '('" actually concludes this case. We could just use "if (i-1>=0 && i-longest[i-1]-1 >=0 && s[i-longest[i-1]-1] == '(')"

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] == ')' && 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);
            }
        }
        return curMax;
    }