标签：bre style ++ default efault turn rmi log inpu
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 };
标签：bre style ++ default efault turn rmi log inpu
原文地址：http://www.cnblogs.com/devin-guwz/p/6507086.html
