2.1.13 Permutation Sequence

Link: https://oj.leetcode.com/problems/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.

这题完全没有思路。再做。

以下代码参考：http://www.lifeincode.net/programming/leetcode-permutation-sequence-java/

Time: O(n), 

But this blog said its O(n^2) because removing elements from array is O(n) //What if removing an element from an ArrayList?

• ArrayList 对于随机位置的add/remove，时间复杂度为 O(n),但是对于列表末尾的添加/删除操作,时间复杂度是 O(1). (Ref: http://blog.csdn.net/renfufei/article/details/17077425)

http://blog.csdn.net/linhuanmars/article/details/22028697

public class Solution {    public String getPermutation(int n, int k) {        int t = 1;        ArrayList<Integer> num = new ArrayList<Integer> ();        for(int i = 1; i <=n; i++){            t *= i;            num.add(i);//num = {1, 2, 3}        }        //t = n!        k--;//from 1-indexed to 0-indexed        t /=n;//t = (n-1)!        StringBuffer sb = new StringBuffer();        for(int i = n-1; i >=1; i--){//i starts with n-1 because t = (n-1)!, need to divide (n-1)            int p = k/t;            int digit = num.get(p);            sb.append(String.valueOf(digit));//note to use "String.valueOf"!            num.remove(p);            k %= t;            t /=i;        }        sb.append(String.valueOf(num.get(0)));        return sb.toString();    }}