20. Valid Parentheses

Posted 2020-08-29 会咬人的兔子

20. Valid Parentheses

• Total Accepted: 167254
• Total Submissions: 517330
• Difficulty: Easy
• Contributors: Admin

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.

Solution:

Use stack. When we meet a left parentheses, we push it into the stack. When we meet a right parentheses, we will return false if the stack is empty, otherwise, we will pop the top of the stack to find if it is the corresponding left parentheses.

 1 class Solution {
 2 public:
 3     bool isValid(string s) {
 4         stack<char> st;
 5         for(int i = 0; i < s.size(); i++){
 6             if(s[i] == ‘(‘ || s[i] == ‘{‘ || s[i] == ‘[‘){
 7                 st.push(s[i]);
 8             }else if(s[i] == ‘)‘ || s[i] == ‘}‘ || s[i] == ‘]‘){
 9                 if(st.empty()){
10                     return false;
11                 }else{//只需判断false的情况，最后再pop就可以了
12                     if(s[i] == ‘)‘ && st.top() != ‘(‘) return false;
13                     if(s[i] == ‘]‘ && st.top() != ‘[‘) return false;
14                     if(s[i] == ‘}‘ && st.top() != ‘{‘) return false;
15                     st.pop();
16                 }
17             }
18         }
19         //return true;此时不能直接return true，因为stack里可能会有剩余的左括号
20         return st.empty();
21     }
22 };