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Commit 84f041e

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add LeetCode 20. 有效的括号
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![](https://imgconvert.csdnimg.cn/aHR0cHM6Ly9jZG4uanNkZWxpdnIubmV0L2doL2Nob2NvbGF0ZTE5OTkvY2RuL2ltZy8yMDIwMDgyODE0NTUyMS5qcGc?x-oss-process=image/format,png)
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>仰望星空的人,不应该被嘲笑
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## 题目描述
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给定一个只包括` '(',')','{','}','[',']' `的字符串,判断字符串是否有效。
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有效字符串需满足:
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左括号必须用相同类型的右括号闭合。
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左括号必须以正确的顺序闭合。
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注意空字符串可被认为是有效字符串。
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示例 1:
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```javascript
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输入: "()"
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输出: true
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```
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示例 2:
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```javascript
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输入: "()[]{}"
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输出: true
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```
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示例 3:
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```javascript
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输入: "(]"
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输出: false
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```
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示例 4:
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```javascript
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输入: "([)]"
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输出: false
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```
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示例 5:
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```javascript
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输入: "{[]}"
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输出: true
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```
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## 题解
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发现越靠后的左括号,最先匹配,也就是`后进先出`的思想,于是考虑使用栈这个数据结构。
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```javascript
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/**
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* @param {string} s
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* @return {boolean}
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*/
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var isValid = function(s) {
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// 如果是奇数,不可能匹配成功,直接返回false
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if(s.length & 1) return false
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let stack = []
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for(let i=0;i<s.length;i++){
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if(s[i] === '(' || s[i] === '{' || s[i] === '[') stack.push(s[i])
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else if(s[i] === ')' && stack[stack.length-1] === '(') stack.pop()
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else if(s[i] === '}' && stack[stack.length-1] === '{') stack.pop()
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else if(s[i] === ']' && stack[stack.length-1] === '[') stack.pop()
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else return false
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}
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return !stack.length
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};
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```
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## 最后
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文章产出不易,还望各位小伙伴们支持一波!
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往期精选:
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<a href="https://github.com/Chocolate1999/Front-end-learning-to-organize-notes">小狮子前端の笔记仓库</a>
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<a href="https://yangchaoyi.vip/">访问超逸の博客</a>,方便小伙伴阅读玩耍~
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![](https://img-blog.csdnimg.cn/2020090211491121.png#pic_center)
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```javascript
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学如逆水行舟,不进则退
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```

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