Leetcode153: 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.

在n!个排列中，第一位的元素总是(n-1)!一组出现的，也就说如果p = k / (n-1)!，那么排列的最开始一个元素一定是nums[p]。
假设有n个元素，第K个permutation是
a1, a2, a3, .....   ..., an
那么a1是哪一个数字呢？
那么这里，我们把a1去掉，那么剩下的permutation为
a2, a3, .... .... an, 共计n-1个元素。 n-1个元素共有(n-1)!组排列，那么这里就可以知道
设变量K1 = K
a1 = K1 / (n-1)!
同理，a2的值可以推导为
a2 = K2 / (n-2)!
K2 = K1 % (n-1)!
 .......
a(n-1) = K(n-1) / 1!
K(n-1) = K(n-2) /2!
an = K(n-1)

class Solution {public:    string getPermutation(int n, int k) {        vector<int> nums(n);          int pCount = 1;          for(int i = 0 ; i < n; ++i)         {              nums[i] = i + 1;              pCount *= (i + 1);          }            k--;          string res = "";          for(int i = 0 ; i < n; i++)         {              pCount = pCount/(n-i);              int selected = k / pCount;              res += ('0' + nums[selected]);              for(int j = selected; j < n-i-1; j++)                  nums[j] = nums[j+1];              k = k % pCount;          }          return res;      }};