20. Valid Parentheses(C++)

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 :

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

better solution:

 1 class Solution {
 2 public:
 3     bool isValid(string s) {
 4         stack<char> paren;
 5         for (char& c : s) {
 6             switch (c) {
 7                 case ‘(‘:
 8                 case ‘{‘:
 9                 case ‘[‘: paren.push(c); break;
10                 case ‘)‘: if (paren.empty() || paren.top()!=‘(‘) return false; else paren.pop(); break;
11                 case ‘}‘: if (paren.empty() || paren.top()!=‘{‘) return false; else paren.pop(); break;
12                 case ‘]‘: if (paren.empty() || paren.top()!=‘[‘) return false; else paren.pop(); break;
13                 default: ; // pass
14             }
15         }
16         return paren.empty() ;
17     }
18 };