【leetcode】Longest Valid Parentheses

最新推荐文章于 2025-11-26 13:52:00 发布

最新推荐文章于 2025-11-26 13:52:00 发布 · 53 阅读

文章标签：

#c/c++

本文探讨如何通过编程解决寻找字符串中最长有效括号子串的问题，详细介绍了两种算法解决方案，包括使用栈和计数的方法，适用于理解并实现括号匹配的逻辑。

Question：

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.

Anwser 1 ：

class Solution {
public:
    int longestValidParentheses(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        stack<int> S;
        
        for(int i = 0; i < s.size(); ++i){      // "("
            if (s[i] == '(') {
                S.push(i);
            } else if (!S.empty()){     //')'
                s[i] = 'k';
                s[S.top()] = 'k';
                S.pop();
            }
        }
        // example: "(()" = "(kk";  "(()()" = "(kkkk"
        
        int maxLength = 0;
        int length = 0;
        for(int i = 0; i < s.size(); ++i){  // count of "k"
            if (s[i] =='k'){
                ++length;                
                if (maxLength < length) {
                    maxLength = length;
                }
            } else {
                length = 0;
            }
        }
        
        return maxLength;
    }
};

Anwser 2 ：

class Solution {
public:
    struct node {
        char c;
        int idx;
        node() {}
        node(char _c, int _idx): c(_c), idx(_idx) {}        
    };
    
    int longestValidParentheses(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        stack<node> st;
        st.push(node(')', -1));
        int res = 0;
        
        for (int i = 0; i < s.size(); ++i)
        {
            char c = s[i];
            if (c == '(') {
                st.push(node(c, i));
            } else {
                node top = st.top();
                if (top.c == '(') {
                    st.pop();
                    res = max(res, i - st.top().idx);   //  count or length
                } else {
                    st.push(node(c, i));
                }
            }
        }
        
        return res;
    }
};