LeetCode——Permutation Sequence，leetcode

LeetCode——Permutation Sequence，leetcode

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.

原題連結：https://oj.leetcode.com/problems/permutation-sequence/

從n個數的全排列中找出第k個排列。

n開頭的排列有(n-1)!個。k/(n-1)!可確定第一個數字，在餘下的(n-1)!中找k%(n-1)!個。

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();}}