Leetcode 20. Valid Parentheses

Posted 2020-08-06 Deribs4

20. Valid Parentheses

• Total Accepted: 128586
• Total Submissions: 418560
• Difficulty: Easy

 

Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the input string is valid.

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

 

思路：利用栈来判断配对的括号，如果都可以配对成功，最后栈的元素数量应该为0。题目给定的字符串只包含括号，所以每次遇到是(‘, ‘{‘或者‘[‘，直接入栈；遇到‘]‘,‘)‘或者‘]‘，只有当栈不为空&&栈顶元素与之配对时，才弹出栈，否则直接返回false。

 

代码：

 1 class Solution {
 2 public:
 3     bool isValid(string s) {
 4         stack<char> cs;
 5         unordered_map<char, char> um;
 6         um[‘(‘] = ‘)‘;
 7         um[‘{‘] = ‘}‘;
 8         um[‘[‘] = ‘]‘;
 9         for (int i = 0; i < s.size(); i++) {
10             if (s[i] == ‘(‘ || s[i] == ‘{‘ || s[i] == ‘[‘) {
11                 cs.push(s[i]);
12                 continue;
13             }
14             if (!cs.empty() && um[cs.top()] == s[i]){
15                 cs.pop();
16                 continue;
17             } 
18             return false;
19         }
20         return cs.empty();
21     }
22 };