LeetCode Permutation Sequence

题目

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 k^th permutation sequence.

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

 

输出从1~n的n个数组成的排列中第K小的。

注意到第i位（从1开始计数）的数字增加1，排列序号增加(n-1)!，可以由此从高位开始依次确定每位的数字。

每次已经被选的数字需要在低位的筛选过程中剔除。

 

代码：

class Solution {public:    string getPermutation(int n, int k) {        string ans;if(n<=0)return ans;int num[9]={1},flag[9]={0};//第i位（从0开始）后面i个数字的排列数，数字i+1是否已经取过num[1]=1;int i,j,id,count;for(i=1;i<9;i++)//获取第i位（从0开始）后面i个数字的排列数的查找表num[i]=num[i-1]*i;for(i=n-1;i>=0;i--)//从头开始确定每位的数字{id=(k-1)/num[i]+1;//确定该位是没取数字中的从小开始的第几个数字count=0;for(j=0;j<n;j++)//寻找该数字{if(flag[j]==0){count++;if(count==id){ans.push_back('1'+j);flag[j]=1;break;}}}k-=(id-1)*num[i];//修改序号，适应下一位}return ans;    }};