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.

題意：給定字符串判斷是否為合法的括號對

思路：利用棧結構，遇到左括號（（、[、{）都壓入棧中，遇到右括號了，若棧為空，則肯定不合法；若棧不為空，則看其是否和棧頂字符向匹配。這裏以最後棧是否為空

 1 class Solution {
 2 public:
 3     bool isValid(string s) 
 4     {
 5         if(s.empty())   return true;
 6         stack<char> stk;
 7         for(int i=0;i<s.size();++i)
 8         {
 9             if(s[i]==‘(‘||s[i]==‘[‘||s[i]==‘{‘)
10                 stk.push(s[i]);
 
11             else 
12             {
13                 if( stk.empty())
14                     return false;
15                  char temp=stk.top();
16                  stk.pop();
17                  if((temp==‘(‘&&s[i] !=‘)‘)||(temp==‘[‘&&s[i] !=‘]‘)||(temp==‘{‘&&s[i] !=‘
 
}‘))
18                      return false;
19             }
20             
21         }
22         return stk.empty();    
23     }
24 };

括號的題還有generate parentheses

[Leetcode] valid parentheses 有效括號對