[LeetCode] Permutation Sequence

 1 class Solution {
 2         vector<bool> m_avaliable;
 3         int m_k;
 4         int m_cnt;
 5         vector<int> m_array;
 6     public:
 7         string m_str;
 8 
 9         void dfs_permutation(int size, int dep)
10         {
11             if(dep == size)
12             {
13                 m_cnt ++;
14                 if(m_cnt == m_k)
15                 {
16                     for(int i = 0; i < size; i++)
17                     {
18                         m_str += m_array[i] + ‘1‘;
19                     }
20                     return;
21                 }
22             }
23 
24             //dfs to the next level
25             // every level has size choices, its time complexity is O(n^n)
26             for(int i = 0; i < size; i++)
27             {
28                 if( m_avaliable[i] == true)
29                 {
30                     //cout <<"dep = " <<dep <<endl;
31                     //cout <<"i   = " <<i <<endl;
32                     m_array[dep] = i;
33                     m_avaliable[i] = false;
34                     dfs_permutation( size, dep+1);
35                     m_avaliable[i] = true;
36                 }
37 
38             }
39 
40 
41 
42         }
43         string getPermutation(int n, int k)
44         {
45             //initinalize the member variable
46             m_avaliable.resize(n, true);
47             m_array.resize(n);
48             m_k = k;
49             m_cnt = 0;
50             m_str.clear();
51 
52             //call dfs
53             dfs_permutation(n, 0);
54 
55             // return 
56             return m_str;
57         }
58 
59 };