Leetcode 32. 最长有效括号

题目描述：给你一个只包含 '(' 和 ')' 的字符串，找出最长有效（格式正确且连续）括号子串的长度。

栈解法：

class Solution {
public:
    
    int longestValidParentheses(string s) {
        stack st;
        int maxLength = 0;
        st.push(-1);
        for(int i = 0; i < s.length(); i++)
        {
            if(s[i] == '(')
            {
                st.push(i);
            }
            else
            {
                st.pop();
                if(st.empty())
                {
                    st.push(i);
                }
                else
                {
                    maxLength = max(maxLength,i-st.top());
                }
            }
            
        }
        return maxLength;
    }
};

动态规划解法：

定义 dp[i] 表示以下标 i 字符结尾的最长有效括号的长度。

(1) s[i] = ')且s[i-1]='('时

dp[i] = dp[i-2]+2

(2) s[i] = ')且s[i-1]=')'时
dp[i]=dp[i−1]+dp[i−dp[i−1]−2]+2

class Solution {
public:
    int longestValidParentheses(string s) {
        int maxans = 0, n = s.length();
        vector dp(n, 0);
        for (int i = 1; i < n; 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 ? dp[i - dp[i - 1] - 2] : 0) + 2;
                }
                maxans = max(maxans, dp[i]);
            }
        }
        return maxans;
    }
};

双指针解法：
 

public:
    int longestValidParentheses(string s) {
        int left = 0, right = 0, maxlength = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s[i] == '(') {
                left++;
            } else {
                right++;
            }
            if (left == right) {
                maxlength = max(maxlength, 2 * right);
            } else if (right > left) {
                left = right = 0;
            }
        }
        left = right = 0;
        for (int i = (int)s.length() - 1; i >= 0; i--) {
            if (s[i] == '(') {
                left++;
            } else {
                right++;
            }
            if (left == right) {
                maxlength = max(maxlength, 2 * left);
            } else if (left > right) {
                left = right = 0;
            }
        }
        return maxlength;
    }
};