60. Permutation Sequence

The set [1,2,3,…,n] contains a total ofn! 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 k^th permutation sequence.

思路: 将元素按从小到大的顺序保存在list中,从第一个开始判断应该是什么数字,依次根据下标从list中选取数据,然后删除已用的数据.

code:

public class Solution {
    public String getPermutation(int n, int k) {
        StringBuilder res=new StringBuilder("");
        int all=1;
        int temp=n-1;
        List<Integer> list=new ArrayList<Integer>();
        for(int i=1;i<=n;i++)
            list.add(i);
        while(temp>1)
            all*=(temp--);
         
        for(int i=0;i<n-1;i++){
            temp=(k-1)/all;// 注意这里是k-1
            res.append(list.get(temp));
            list.remove(temp);
            k-=temp*all;
            all/=(n-i-1);
        }
        res.append(list.get(0));
        return res.toString();
    }
}