【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):

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.

 class Solution {
 public:
     string getPermutation(int n, int k) {
         string result;
         bool used[];
         memset(used, false, );
         int base[];
         base[] = ;
         for (int i = ; i <= n; ++i) {
             base[i] = base[i - ] * i;
         }
         --k;
         for (int i = ; i < n; ++i) {
             int p = k / base[n -  - i];
             for (int j = ; j <= n; ++j) {
                 if (!used[j] && p-- == ) {
                     result.push_back('' + j);
                     used[j] = true;
                     break;
                 }
             }
             k %= base[n -  - i];
         }
         return result;
     }
 };

n个数组的全排列中，以1开头的个数(n-1)!个，同样以2..n开头的个数有(n-1)!个，这样根据(k-1)/[(n-1)!]可知第一个数是多少，依次类推...