leetcode 32:最长有效括号

一、

先对字符串进行遍历  首先从前往后排除不配对的')'

首次遍历后的字符串被分成若干个字符串  再对这些字符串 从后往前排除不配对的'('

int longestValidParentheses(std::string s) {
    std::stack<int> ss1;
    std::stack<char> ss2;
    std::stack<char> ss3;
    std::vector<int> q1;
    std::vector<int> q2;
    int len = s.size();
    int n = 0;
    int max2 = 0;
    int t1 = INT_MAX;
    int t2 = 0;
    int s1 = 0;
    int s2 = 0;
    for (int j = 0; j < len; j++) {
        if (s[j] == '(') {
            ss1.push(j);
            ss2.push(s[j]);
        } else if (s[j] == ')') {
            if (!ss2.empty()) {
                n = n + 2;
                if (t1 > ss1.top())
                    t1 = ss1.top();
                t2 = j;
                ss1.pop();
                ss2.pop();
            } else {
                if (max2 < n) {
                    max2 = n;
                    s1 = t1;
                    s2 = t2;
                    q1.push_back(s1);
                    q2.push_back(s2);
                }
                t1 = j + 1;
                n = 0;
                continue;
            }
        }
    }
    if (max2 < n) {
        s1 = t1;
        s2 = t2;
        q1.push_back(s1);
        q2.push_back(s2);
    }
    max2 = 0;
    while (!q1.empty()) {
        int s1 = q1.back();
        q1.pop_back();
        int s2 = q2.back();
        q2.pop_back();
        n = 0;
        for (int j = s2; j >= s1; j--) {
            if (s[j] == ')') {
                ss3.push(s[j]);
            } else if (s[j] == '(') {
                if (!ss3.empty()) {
                    n = n + 2;
                    ss3.pop();
                } else {
                    if (max2 < n) {
                        max2 = n;
                    }
                    n = 0;
                    continue;
                }
            }
        }
        if (max2 < n)
            max2 = n;
    }
    return max2;
}

二、动态规划

int longestValidParentheses(std::string s) {
    std::stack<int>sta;
    int max=0;
    for(int i=0;i<s.size();i++){
        if(s[i]==')'&&!sta.empty()&&s[sta.top()]=='('){
            sta.pop();
            if(sta.empty()){
                max=i+1;
            }
            else if(max<i-sta.top())
            {
                max=i-sta.top();
            }
        }else
        {
            sta.push(i);
        }
    }
    return max;
}