32. Longest Valid Parentheses

https://leetcode.com/problems/longest-valid-parentheses/

C++

class Solution {
public:
    int longestValidParentheses(string s) {
        int result = 0;
        stack<int> stk;

        for (int i = 0; i < s.length(); i++) {
            if (s[i] == ')' && !stk.empty() && s[stk.top()] == '(') {
                stk.pop();
                if (stk.empty()) {
                    result = i + 1;
                } else {
                    result = max(result, i - stk.top());
                }
            } else {
                stk.push(i);
            }
        }

        return result;
    }
};

Go

func longestValidParentheses(s string) int {
    result := 0
    stack := []int{}

    for i, c := range s {
        if c == ')' && len(stack) > 0 && s[stack[len(stack) - 1]] == '(' {
            stack = stack[0:len(stack)-1]
            if len(stack) > 0 {
                if i - stack[len(stack)-1] > result {
                    result = i - stack[len(stack)-1]
                }
            } else {
                result = i + 1
            }
        } else {
            stack = append(stack, i)
        }
    }

    return result
}

JavaScript

/**
 * @param {string} s
 * @return {number}
 */
var longestValidParentheses = function(s) {
  var result = 0
  var stack = []

  for (let i = 0; i < s.length; i++) {
    if (s[i] === ')' && stack.length && s[stack[stack.length - 1]] === '(') {
      stack.pop()
      if (stack.length) {
        result = Math.max(result, i - stack[stack.length - 1])
      } else {
        result = i + 1
      }
    } else {
      stack.push(i)
    }
  }

  return result
}

Python

class Solution(object):

    def longestValidParentheses(self, s):
        """
        :type s: str
        :rtype: int
        """
        result, stack = 0, []
        for i, c in enumerate(s):
            if c == ')' and len(stack) > 0 and s[stack[-1]] == '(':
                stack.pop()
                if len(stack) > 0:
                    result = max(result, i - stack[-1])
                else:
                    result = i + 1
            else:
                stack.append(i)
        return result