LeetCode——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.

原题链接：https://oj.leetcode.com/problems/permutation-sequence/

从n个数的全排列中找出第k个排列。

n开头的排列有(n-1)!个。k/(n-1)!可确定第一个数字，在余下的(n-1)!中找k%(n-1)!个。

public class PermutationSequence {
public String getPermutation(int n, int k) {
List<Integer> list = new ArrayList<Integer>();
for(int i=1;i<=n;i++)
list.add(i);
int mod = 1;
for (int i = 1; i <= n; i++) {
mod = mod * i;
}
k--;
StringBuilder builder = new StringBuilder();
for(int i=0;i<n;i++){
mod /= (n-i);
int index = k / mod;
k %= mod;
builder.append(list.get(index));
list.remove(index);
}
return builder.toString();
}
}