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 kth permutation sequence.

思路：

没有呢还。。。

string getPermutation(int n, int k)    {        vector
     
      factorial(n + 1, 1);//保留阶乘        for (int i = 1; i <= n;i++)                 factorial[i] = i * factorial[i - 1];        vector
      
        digits = { '1', '2', '3', '4', '5', '6', '7', '8', '9' };        string result;        int num = n - 1;        while (num)        {            int t = (k-1) / factorial[num];   //是k/(n-1)! 计算第一位数字            k = k - t * factorial[num];            result.push_back(digits[t]);            digits.erase(digits.begin() + t);            num--;        }        result.push_back(digits[k - 1]);        return result;            }
      
     

 

参考：