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.

 

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    这道题可以直接穷举然后数第k 个，或者 迭代k次得到返回值，或者直接计算出结果。我用是直接计算结果，很多细节上需要处理。

 

#include 
      
       #include 
       
        using namespace std;class Solution {public:    string getPermutation(int n, int k) {        if(n <2)    return "1";        int tab[10]={
    1};        for(int i =1;i<=9;i++)            tab[i] = tab[i-1]*i;        int remain[9]={
    0};        for(int i=0;i<9;i++)            remain[i]=i+1;        int cnt = n;        string ret = "";        while(cnt>0){            cnt--;            int num = (k-1)/tab[cnt];//            cout<<"num="<
        
         <