Leetcode sorting: k-th sorting

I forgot to write the next arrangement in STL.

It takes more than one hour to write data. If recursive writing is used, the error will be saved in a list, and then K/(n-1) will be used )! The factorial is the number to be deleted, but after observation,

For example, list = {1, 2, 3}

Divided into three groups:

1 {2, 3}

2 {1, 3}

3 {1, 2}

Determine the group in which nyoj 511 is located, and then determine the number of nyoj in which group.

Find 3rd permutation, 3% 2 = 1, delete list is 3rd number 3, in fact it is 2nd tree 2, so the calculation method is (k-1)/(n-1 )!

For the next group, K % (n-1 )! No. 4th, 4%! = 0. It should actually be the second 2.

This idea is similar to nyoj's ball fall (nyoj 511)

 1 public class Solution { 2      3     private String ans=""; 4     public int calu(int n) 5     { 6         if(n==0) return 1; 7         int sum=1; 8         for(int i=2;i<=n;i++) 9         {10             sum*=i;11         }12         return sum;13     }14     15     public String getPermutation(int n, int k) {16         ArrayList<Integer> arry=new ArrayList<Integer>();17         for(int i=1;i<=n;i++)18         {19             arry.add(i);20         }21         22         get(k,calu(n-1),arry);23         return ans;24         25     }26     public void get(int k,int n1,ArrayList<Integer> list)27     {28         if(list.size()==1)29         {30             ans+=list.remove(0);31             return;32         }33            int a=list.remove((k-1)/n1);34            ans+=a;35            int te=k%n1;36            if(te==0) te=n1;37         get(te,n1/list.size(),list);38         39         40     }41 }

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