LeetCode.20:Valid Parentheses

最新推荐文章于 2022-09-01 00:56:47 发布

RyanWang0000

最新推荐文章于 2022-09-01 00:56:47 发布

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分类专栏： LeetCode 文章标签： LeetCode

本文链接：https://blog.youkuaiyun.com/qq_32350719/article/details/90183151

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本文介绍了一个使用栈数据结构来验证括号是否正确匹配的算法。该算法能够处理圆括号、方括号和花括号，并确保它们成对出现且顺序正确。通过一系列示例，展示了如何判断字符串中的括号序列是否有效。

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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

思路：简单题，运用栈的后进先出特性去匹配即可，注意要排除各种意外情况

package com.Easy.Ryan;

import java.util.Stack;

public class ValidParentheses {

	public static void main(String[] args) {
		ValidParentheses validParentheses=new ValidParentheses();
		String string=  "{{)}";
		System.out.println(validParentheses.isValid(string));
	}
	public boolean isValid(String s) {
		boolean result = false;
		Stack<Character> stack=new Stack<>();
		Character character;
		if (s.length()%2==1) {
			return false;
		}
		if (s.length()==0) {
			return true;
		}
		for (int i = 0; i < s.length(); i++) {
			if (s.charAt(i)=='(' || s.charAt(i)=='[' || s.charAt(i)=='{') {
				stack.push(s.charAt(i));
			}
			else  {
				if (stack.isEmpty()) {
					return false;
				}
				character=stack.pop();
				if (s.charAt(i)==')'&& character=='(' || s.charAt(i)==']'&& character=='[' || s.charAt(i)=='}'&& character=='{') {
					result=true;
				}
				else {
					return false;
				}
			}
		}
		if (!stack.isEmpty()) {
			result=false;
		}
		
		return result;
    }
}