Valid Parentheses-优快云博客

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/87625737

本文介绍了一个LeetCode上的经典算法问题——有效括号的解决方案。该问题要求判断输入字符串中的括号（包括圆括号、方括号和花括号）是否正确配对并闭合。文章提供了详细的解题思路及AC代码实现。

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

题目

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true
Example 2:

Input: "()[]{}"
Output: true
Example 3:

Input: "(]"
Output: false
Example 4:

Input: "([)]"
Output: false
Example 5:

Input: "{[]}"
Output: true

题意：判断输入的字符串(由三种括号组成)，是否满足括号的规范。
AC code

class Solution {
    public boolean isValid(String s) {
    	
    	
    	String str="";
    	for (int i = 0; i < s.length(); i++) {
			char ch=s.charAt(i);
    		if(ch=='(') {
    			str+=")";
    		}else if(ch=='[') {
    			str+="]";
    		}else if(ch=='{') {
    			str+="}";
    		}else if(ch==')'&&str.length()!=0) {//right
    			if(str.charAt(str.length()-1)!=ch)
    				return false;
    			str=str.substring(0, str.length()-1);
    		}else if(ch==']'&&str.length()!=0) {//right
    			if(str.charAt(str.length()-1)!=ch)
    				return false;
    			str=str.substring(0, str.length()-1);
    		}else if(ch=='}'&&str.length()!=0) {//right
    			if(str.charAt(str.length()-1)!=ch)
    				return false;
    			str=str.substring(0, str.length()-1);
    		}
    		else {
    			return false;
    		}
		}
    	if(str.length()!=0)
    		return false;
    	return true;
    }
}