60. 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.

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第一个数决定在第几个[0,(n-1)!],[(n-1)!,2(n-1)!]……这样的区间内，第一个确定之后确定第二个……以此类推

public class Solution {    public String getPermutation(int n, int k) {        String ret = "";        int s = 1;        for (int i = n-1;i>1;i--){            s*=i;        }        List <Integer> ku = new LinkedList <>();        for(int i = 1;i<=n;i++){            ku.add(i);        }        k--;        for(int i = 1;i<=n;i++){            int tem = k/s;            k = k%s;            if(n-i!=0)s = s/(n-i);            ret+=""+ku.get(tem);            ku.remove(tem);        }        return ret;    }}