【leetcode系列】20.有效的括号

 

题目描述

1Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
 2
 3An input string is valid if:
 4
 5Open brackets must be closed by the same type of brackets.
 6Open brackets must be closed in the correct order.
 7Note that an empty string is also considered valid.
 8
 9Example 1:
10
11Input: "()"
12Output: true
13Example 2:
14
15Input: "()[]{}"
16Output: true
17Example 3:
18
19Input: "(]"
20Output: false
21Example 4:
22
23Input: "([)]"
24Output: false
25Example 5:
26
27Input: "{[]}"
28Output: true

思路

关于这道题的思路，邓俊辉讲的很是好，没有看过的同窗能够看一下, 视频地址。

使用栈,遍历输入字符串

若是当前字符为左半边括号时，则将其压入栈中

若是遇到右半边括号时，分类讨论：

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

2）若此时栈为空，则直接返回false

3）若不为对应的左半边括号，反之返回false

 

值得注意的是，若是题目要求只有一种括号，那么咱们其实能够使用更简洁，更省内存的方式 - 计数器来进行求解，而
没必要要使用栈。

事实上，这类问题还能够进一步扩展，咱们能够去解析相似HTML等标记语法， 好比

 

 

关键点解析

1. 栈的基本特色和操做

2. 若是你用的是JS没有现成的栈，能够用数组来模拟
   入：push 出: pop

入：push 出 shift 就是队列

代码

• 语言支持：JS，Python

Javascript Code:

1/*
 2 * @lc app=leetcode id=20 lang=javascript
 3 *
 4 * [20] Valid Parentheses
 5 *
 6 * https://leetcode.com/problems/valid-parentheses/description/
 7 *
 8 * algorithms
 9 * Easy (35.97%)
10 * Total Accepted:    530.2K
11 * Total Submissions: 1.5M
12 * Testcase Example:  '"()"'
13 *
14 * Given a string containing just the characters '(', ')', '{', '}', '[' and
15 * ']', determine if the input string is valid.
16 * 
17 * An input string is valid if:
18 * 
19 * 
20 * Open brackets must be closed by the same type of brackets.
21 * Open brackets must be closed in the correct order.
22 * 
23 * 
24 * Note that an empty string is also considered valid.
25 * 
26 * Example 1:
27 * 
28 * 
29 * Input: "()"
30 * Output: true
31 * 
32 * 
33 * Example 2:
34 * 
35 * 
36 * Input: "()[]{}"
37 * Output: true
38 * 
39 * 
40 * Example 3:
41 * 
42 * 
43 * Input: "(]"
44 * Output: false
45 * 
46 * 
47 * Example 4:
48 * 
49 * 
50 * Input: "([)]"
51 * Output: false
52 * 
53 * 
54 * Example 5:
55 * 
56 * 
57 * Input: "{[]}"
58 * Output: true
59 * 
60 * 
61 */
62/**
63 * @param {string} s
64 * @return {boolean}
65 */
66var isValid = function(s) {
67    let valid = true;
68    const stack = [];
69    const mapper = {
70        '{': "}",
71        "[": "]",
72        "(": ")"
73    }
74
75    for(let i in s) {
76        const v = s[i];
77        if (['(', '[', '{'].indexOf(v) > -1) {
78            stack.push(v);
79        } else {
80            const peak = stack.pop();
81            if (v !== mapper[peak]) {
82                return false;
83            }
84        }
85    }
86
87    if (stack.length > 0) return false;
88
89    return valid;
90};

Python Code:

1    class Solution:
 2        def isValid(self,s):
 3          stack = []
 4          map = {
 5            "{":"}",
 6            "[":"]",
 7            "(":")"
 8          }
 9          for x in s:
10            if x in map:
11              stack.append(map[x])
12            else:
13              if len(stack)!=0:
14                top_element = stack.pop()
15                if x != top_element:
16                  return False
17                else:
18                  continue
19              else:
20                return False
21          return len(stack) == 0

扩展

若是让你检查XML标签是否闭合如何检查， 更进一步若是要你实现一个简单的XML的解析器，应该怎么实现？