LeetCode 32 括号匹配

标签：character   依次   思路   字符   bst   idp   遍历   rac   出栈   

[LeetCode 32] Longest Valid Parentheses

题目

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

测试案例

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

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

思路

1. 采用栈数据结构。栈中存放各字符的下标。初始时里面放入-1。

2. 从左至右依次遍历每个字符。当前字符为左括号就进栈。当前字符为右括号时，如果栈中存在左括号，则出栈。否则，入栈。
3. 每当都元素，记下标为 i ，进栈时，就用 i - stack.peek() - 1 更新 max。

4. 遍历结束后，需要用 n - stack.peek() - 1 更新 max。

代码如下

class Solution {
    public int longestValidParentheses(String s) {        
        int max = 0, n = s.length(), temp, index = 0; 
        if(n == 0){
            return 0;
        }
        int[] stack = new int[n + 1];        
        stack[index++] = -1;
        for(int i = 0; i < n; i++){
            if(s.charAt(i) == ‘(‘ || (temp = stack[index - 1]) == -1 || 
               s.charAt(temp) == ‘)‘){                               
                if((temp = i - stack[index - 1] - 1) > max){
                    max = temp;
                }                
                stack[index++] = i;
            }
            else{
                index--;                
            }            
        }
        if((temp = n - stack[index - 1] - 1) > max){
            max = temp;
        }        
        return max;
    }
}

LeetCode 32 括号匹配

标签：character   依次   思路   字符   bst   idp   遍历   rac   出栈   

原文地址：https://www.cnblogs.com/echie/p/9589097.html