LeetCode : Longest Valid Parentheses [java]

最新推荐文章于 2020-04-23 13:12:43 发布

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#LeetCode #Longest Valid Parent

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本文介绍了一种使用栈和标记数组的方法来寻找给定字符串中最长的有效括号子串长度。通过遍历字符串并将左括号的位置存入栈中，当遇到右括号时检查栈是否为空来确定是否构成有效括号对，并使用标记数组记录有效的括号位置。

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.

思路：使用另一个等长数组标记合法的连续区间，之后扫描最大的连续区。

import java.util.HashMap;
import java.util.Map;
import java.util.Stack;

public class Solution {
    public int longestValidParentheses(String s) {
		Stack<Map<Character,Integer>> stack = new Stack<>();
		boolean[] flag = new boolean[s.length()];
		if (s == null || s.length() == 0) {
			return 0;
		}
		for (int i = 0; i < s.length(); i++) {
			char ch = s.charAt(i);
			if (ch == '(') {
				Map<Character,Integer> t_map = new HashMap<Character, Integer>();
				t_map.put(ch, i);
				stack.push(t_map);
			} else if (ch == ')' && !stack.isEmpty()) {
				flag[i] = true;
				Map<Character,Integer> t_map = stack.pop();
				flag[t_map.get('(')] = true;
			}
		}
		int curLength = 0;
		int maxLength = 0;
		for (int i = 0; i < flag.length; i++) {
			if (flag[i]) {
				curLength++;
			} else {
				curLength = 0;
			}
			maxLength = maxLength > curLength ? maxLength : curLength;
		}
		return maxLength;
    }
    
    
}