https://leetcode.com/problems/valid-parentheses/description
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
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。
使用栈,遍历输入字符串
如果当前字符为左半边括号时,则将其压入栈中
如果遇到右半边括号时,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
(图片来自: https://github.com/MisterBooo/LeetCodeAnimation)
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 不必要使用栈。
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
- 栈的基本特点和操作
- 如果你用的是JS没有现成的栈,可以用数组来模拟 入: push 出: pop
入: push 出 shift 就是队列
- 语言支持:JS,Python
Javascript Code:
/*
* @lc app=leetcode id=20 lang=javascript
*
* [20] Valid Parentheses
*
* https://leetcode.com/problems/valid-parentheses/description/
*
* algorithms
* Easy (35.97%)
* Total Accepted: 530.2K
* Total Submissions: 1.5M
* Testcase Example: '"()"'
*
* 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
*
*
*/
/**
* @param {string} s
* @return {boolean}
*/
var isValid = function(s) {
let valid = true;
const stack = [];
const mapper = {
'{': "}",
"[": "]",
"(": ")"
}
for(let i in s) {
const v = s[i];
if (['(', '[', '{'].indexOf(v) > -1) {
stack.push(v);
} else {
const peak = stack.pop();
if (v !== mapper[peak]) {
return false;
}
}
}
if (stack.length > 0) return false;
return valid;
};
Python Code:
class Solution:
def isValid(self,s):
stack = []
map = {
"{":"}",
"[":"]",
"(":")"
}
for x in s:
if x in map:
stack.append(map[x])
else:
if len(stack)!=0:
top_element = stack.pop()
if x != top_element:
return False
else:
continue
else:
return False
return len(stack) == 0
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?