[LeetCode] Permutation Sequence

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):

"123"
"132"
"213"
"231"
"312"
"321"

Given n and k, return the k^th permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

DFS从小到大枚举排列，到K时输出。

 1 class Solution {
 2 private:
 3     int a[10];
 4     bool canUse[10];
 5     string ret;
 6 public:
 7     void dfs(int dep, int maxDep, int &k)
 8     {
 9         if (k == 0)
10             return;
11             
12         if (dep == maxDep)
13         {
14             k--;
15             if (k == 0)
16             {
17                 ret = "";
18                 for(int i = 0; i < maxDep; i++)
19                     ret += (char)(a[i] + '0');
20                 return;
21             }
22         }
23         
24         for(int i = 1; i <= maxDep; i++)
25             if (canUse[i])
26             {
27                 canUse[i] = false;
28                 a[dep] = i;
29                 dfs(dep + 1, maxDep, k);
30                 canUse[i] = true;
31             }
32     }
33     
34     string getPermutation(int n, int k) {
35         // Start typing your C/C++ solution below
36         // DO NOT write int main() function
37         memset(canUse, true, sizeof(canUse));
38         
39         dfs(0, n, k);
40         
41         return ret;
42     }
43 };