60. Permutation Sequence

Posted 2020-08-13 Machelsky

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.

参考解析：http://bangbingsyb.blogspot.com/2014/11/leetcode-permutation-sequence.html

思路：先算n!。.然后k/(n-1)!＋1则为现在的第一个数字。更新k%=(n-1)!, n=n-1.

public class Solution {

    public String getPermutation(int n, int k) {
        if(n<=0)
        {
            return "";
        }
        List<Integer> list=new ArrayList<Integer>();
        int fac=1;
        
        for(int i=1;i<=n;i++)
        {
            list.add(i);
            fac*=i;
        }
        
        k--; //zero based
        
        StringBuilder sb=new StringBuilder();
        
        while(n>0)
        {
            fac/=n;
            sb.append(list.remove(k/fac));  //di ji zu
            
            k%=fac; //di ji ge 
            
            n--;
        }
        return sb.toString();
    }
}