LeetCode刷题笔录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):

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.

开始的想法是DFS，按顺序得到所有的permutation，直到得到第K个为止。

public class Solution {public String getPermutation(int n, int k) {boolean[] used = new boolean[n];List<String> res = new ArrayList<String>();helper(0, new int[] { 1 }, k, n, used, new StringBuffer(), res);return res.get(0);}public void helper(int level, int[] count, int k, int n, boolean[] used,StringBuffer str, List<String> res) {if (level == n) {if (count[0] < k) {count[0]++;return;} else {res.add(str.toString());return;}}if (count[0] <= k) {for (int i = 1; i <= n; i++) {if (used[i - 1] == true)continue;used[i - 1] = true;str.append(i);helper(level + 1, count, k, n, used, str, res);used[i - 1] = false;str.deleteCharAt(str.length() - 1);}}}}

在自己机器上没什么问题，不过在LeetCode的OJ里总会超时。这样的时间复杂度确实有点太高了。