28 字符串的排列

 1 class Solution {
 2 private:
 3     vector<string> res ;
 4 public:
 5     vector<string> Permutation(string str) {
 6         if (str.empty())
 7             return res ;
 8         sort(str.begin() , str.end()) ;
 9         bool isUsed[15]  = {false};
10         string temp = "" ;
11         backtracking(str,isUsed,temp) ;
12         return res ;
13     }
14     
15     void backtracking(string str , bool isUsed[] , string temp){
16         if (temp.size() == str.size()){
17             res.push_back(temp) ;
18             return ;
19         }
20         for(int i = 0 ; i < str.size() ; i++){
21             if (isUsed[i])
22                 continue ;
23             //处理重复字符  比如两个bac1c2  bac2c1 只算其中一种
24             if (i!=0 && str[i] == str[i-1] && !isUsed[i-1])
25                 continue ;
26             isUsed[i] = true ;
27             temp += str[i] ;
28             backtracking(str,isUsed,temp) ;
29             temp.pop_back();
30             isUsed[i] = false ;
31         }
32     }
33 };