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):

Given n and k, return the kth permutation sequence.

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

标签 Backtracking Math

类似题目 (M) Next Permutation (M) Permutations

思路：

1.对于由{1,2,...,n}组成的第k个数（k从0开始），k = a * （n-1）! + b，其中a表示的是{1,2,...,n}中第a个数，b表示的是由{1,2,...,n}去掉第a个数后剩下的数（维持升序排列）构成的数组成的排列的第b个排列。

2.k = b，重复1共n次。

代码：

 1 public class Solution {
 2     public String getPermutation(int n, int k) {
 3         List<Integer> numbers = new ArrayList<>();
 4         int[] factor = new int[n];
 5         int i, t;
 6         String res = "";
 7         factor[0] = 1;
 8         for (i = 1; i < n; ++i)    factor[i] = factor[i - 1] * i;
 9         for (i = 1; i <= n; ++i) numbers.add(i);
10         k--;
11         for (i = n - 1; i >= 0 ; --i) {
12             t = k / factor[i];
13             res = res + numbers.get(t);
14             k = k % factor[i];
15             numbers.remove(t);
16         }
17         return res;
18     }
19 }