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):

  1. "123"
  2. "132"
  3. "213"
  4. "231"
  5. "312"
  6. "321"

 

Given n and k, return the kth permutation sequence.

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

代码：

 1     string getPermutation(int n, int k) {
 2         // IMPORTANT: Please reset any member data you declared, as
 3         // the same Solution instance will be reused for each test case.
 4         int fac[10];
 5         fac[0] = 1;
 6         for(int i = 1; i <= 9; i++){
 7             fac[i] = fac[i-1]*i;
 8         }
 9         string result = "";
10         vector<int> nums;
11         for(int i = 0; i < n; i++)
12             nums.push_back(i+1);
13         for(int i = n; i >= 1; i--){
14             if(k <= fac[i-1]){
15                 result += (nums[0]+'0');
16                 nums.erase(nums.begin());
17             }
18             else if(k == fac[i]){
19                 for(vector<int>::iterator it = nums.end()-1; it >= nums.begin(); it--){
20                     result += (*it+'0');
21                 }
22                 break;
23             }
24             else{
25                 int j = (k-1)/fac[i-1]+1;
26                 k = k - (j-1)*fac[i-1];
27                 vector<int>::iterator it = nums.begin();
28                 while(j > 1){
29                     it++;
30                     j--;
31                 }
32                 result += (*it+'0');
33                 nums.erase(it);
34             }
35         }
36         return result;
37     }

 

posted on 2013-11-14 20:02  waruzhi  阅读(134)  评论(0)  编辑  收藏  举报

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