[LeetCode] 32. Longest Valid Parentheses 最长有效括号

Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.

Example 1:

Input: "(()"
Output: 2
Explanation: The longest valid parentheses substring is "()"

Example 2:

Input: ")()())"
Output: 4
Explanation: The longest valid parentheses substring is "()()"

可以用 DP或者Stack来解。

栈 Stack：定义个start变量来记录合法括号串的起始位置，遍历字符串，如果遇到左括号，则将当前下标压入栈，如果遇到右括号，如果当前栈为空，则将下一个坐标位置记录到start，如果栈不为空，则将栈顶元素取出，此时若栈为空，则更新结果和i - start + 1中的较大值，否则更新结果和i - 栈顶元素中的较大值。

Java:

public static int longestValidParentheses(String s) {
	Stack<int[]> stack = new Stack<int[]>();
	int result = 0;
 
	for(int i=0; i<=s.length()-1; i++){
		char c = s.charAt(i);
		if(c=='('){
			int[] a = {i,0};
			stack.push(a);
		}else{
			if(stack.empty()||stack.peek()[1]==1){
				int[] a = {i,1};
				stack.push(a);
			}else{
				stack.pop();
				int currentLen=0;
				if(stack.empty()){
					currentLen = i+1;
				}else{
					currentLen = i-stack.peek()[0];
				}
				result = Math.max(result, currentLen);
			}
		}
	}
 
	return result;
}

Python:

# Time:  O(n)
# Space: O(n)
class Solution2:
    # @param s, a string
    # @return an integer
    def longestValidParentheses(self, s):
        longest, last, indices = 0, -1, []
        for i in xrange(len(s)):
            if s[i] == '(':
                indices.append(i)
            elif not indices:
                last = i
            else:
                indices.pop()
                if not indices:
                    longest = max(longest, i - last)
                else:
                    longest = max(longest, i - indices[-1])
        return longest

C++:

class Solution {
public:
    int longestValidParentheses(string s) {
        int res = 0, start = 0;
        stack<int> m;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] == '(') m.push(i);
            else if (s[i] == ')') {
                if (m.empty()) start = i + 1;
                else {
                    m.pop();
                    res = m.empty() ? max(res, i - start + 1) : max(res, i - m.top());
                }
            }
        }
        return res;
    }
};

　　

类似题目：

[LeetCode] 20. Valid Parentheses 合法括号

[LeetCode] 22. Generate Parentheses

[LeetCode] 241. Different Ways to Add Parentheses

[LeetCode] 301. Remove Invalid Parentheses

　　

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