132 lines
3.1 KiB
Markdown
132 lines
3.1 KiB
Markdown
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## 题目地址
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https://leetcode.com/problems/valid-parentheses/description
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## 题目描述
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```
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Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
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An input string is valid if:
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Open brackets must be closed by the same type of brackets.
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Open brackets must be closed in the correct order.
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Note that an empty string is also considered valid.
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Example 1:
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Input: "()"
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Output: true
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Example 2:
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Input: "()[]{}"
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Output: true
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Example 3:
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Input: "(]"
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Output: false
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Example 4:
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Input: "([)]"
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Output: false
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Example 5:
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Input: "{[]}"
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Output: true
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```
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## 思路
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关于这道题的思路，邓俊辉讲的非常好，没有看过的同学可以看一下，[视频地址](http://www.xuetangx.com/courses/course-v1:TsinghuaX+30240184+sp/courseware/ad1a23c053df4501a3facd66ef6ccfa9/8d6f450e7f7a445098ae1d507fda80f6/)。
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使用栈，遍历输入字符串
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如果当前字符为左半边括号时，则将其压入栈中
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如果遇到右半边括号时，分类讨论：
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1）如栈不为空且为对应的左半边括号，则取出栈顶元素，继续循环
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2）若此时栈为空，则直接返回 false
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3）若不为对应的左半边括号，反之返回 false
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（图片来自： https://github.com/MisterBooo/LeetCodeAnimation)
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> 值得注意的是，如果题目要求只有一种括号，那么我们其实可以使用更简洁，更省内存的方式 - 计数器来进行求解，而
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不必要使用栈。
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> 事实上，这类问题还可以进一步扩展，我们可以去解析类似 HTML 等标记语法， 比如 <p></p> <body></body>
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## 关键点解析
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1. 栈的基本特点和操作
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2. 如果你用的是 JS 没有现成的栈，可以用数组来模拟
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入： push 出：pop
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> 入： push 出 shift 就是队列
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## 代码
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* 语言支持：JS，Python
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Javascript Code:
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```js
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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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let valid = true;
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const stack = [];
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const mapper = {
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'{': "}",
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"[": "]",
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"(": ")"
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}
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for(let i in s) {
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const v = s[i];
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if (['(', '[', '{'].indexOf(v) > -1) {
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stack.push(v);
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} else {
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const peak = stack.pop();
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if (v !== mapper[peak]) {
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return false;
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}
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}
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}
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if (stack.length > 0) return false;
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return valid;
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};
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```
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Python Code:
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```
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class Solution:
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def isValid(self,s):
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stack = []
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map = {
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"{":"}",
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"[":"]",
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"(":")"
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}
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for x in s:
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if x in map:
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stack.append(map[x])
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else:
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if len(stack)!=0:
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top_element = stack.pop()
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if x != top_element:
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return False
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else:
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continue
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else:
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return False
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return len(stack) == 0
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```
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## 扩展
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如果让你检查 XML 标签是否闭合如何检查， 更进一步如果要你实现一个简单的 XML 的解析器，应该怎么实现？
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