leetcode32 最长游戏括号 dp

有一说一，我觉得这题没有到困难级

要保存之前的状态，感觉是很明显的dp

思路和题解一样

class Solution {
public:
    int longestValidParentheses(string s) {
        int len=s.length();
        int ret = 0;
        int *dp=new int[len];
        for(int i=0;i<len;i++)
            dp[i]=0;

        for (int i = 1; i < s.length(); i++) {
            if (s[i] == ')') {
                if (s[i - 1] == '(') {
                    dp[i] = (i >= 2 ? dp[i - 2] : 0) + 2;
                }
                else if (i - dp[i - 1] > 0 && s[i - dp[i - 1] - 1] == '(') {
                    dp[i] = dp[i - 1] + ((i - dp[i - 1]-2) >= 0 ? dp[i - dp[i - 1] - 2] : 0) + 2;
                }
                ret=max(ret,dp[i]);
            }
        }
        return ret;
    }
};

然后看了题解，因为只有( H和)，只需要用left，right记录即可

扫描两遍的原因是 (() 避免这种情况

class Solution {
public:
    int longestValidParentheses(string s) {
        int res = 0;
        int left = 0;
        int mark = 0;
        for (int i = 0; i < s.size(); ++i) {
            int prev_mark = mark;
            mark = max(0, mark + ((s[i] == '(') ? 1 : -1));
            if (mark == 0) {
                if (prev_mark > 0) {
                    res = max(i - left + 1, res);
                } else {
                    left = i + 1;
                }
            }
        }
        mark = 0;
        int right = s.size() - 1;
        for (int i = s.size() - 1; i >= 0; --i) {
            int prev_mark = mark;
            mark = max(0, mark + ((s[i] == ')') ? 1 : -1));
            if (mark == 0) {
                if (prev_mark > 0) {
                    res = max(right - i + 1, res);
                } else {
                    right = i - 1;
                }
            }
        }
        return res;
    }
};