Leetcode 60. 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 kth permutation sequence.

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

s思路：
1. 把所有1-9的排列数预先计算出来，即：{0,1,2,6,24,120,720,5040,40320,362880}。例如：n=4时：

1,2,3,41,2,4,31,3,2,41,3,4,21,4,2,31,4,3,22,1,3,42,1,4,32,3,1,42,3,4,12,4,1,32,4,3,13,1,2,43,1,4,2...

s上面列举了前14个排列。可以看出规律，第一位为１的有６个，为２、３、４的都有６个，为什么是６？因为第三个数的排列数刚好是６。所以，先用k/6得到第一位的数，然后k=k%6,在用k/2得到第二位数，然后更新k=k%2;最后k/1得到第三位数，剩下的数就是第四位数

class Solution {public:    string getPermutation(int n, int k) {        //        //vector<int> fact{0,1,2,6,24,120,720,5040,40320,362880};        int fact[10]={0,1,2,6,24,120,720,5040,40320,362880};//优化：用array比vector速度快！！        //这样不好set<int> ss{0,1,2,3,4,5,6,7,8,9};//        string num="123456789";//简洁的存储数据，不用搞成set<int>,再取出数to_string()，就太不简洁        string res="";        k--;        n--;        while(n>0){            int idx=k/fact[n];            res+=num[idx];            k=k%fact[n];            num.erase(idx,1);//idx开头长度为１            n--;        }        res+=num[0];        return res;    }};