题解 | #环形链表的约瑟夫问题#

用list模拟离开回环的情况，直到剩下最后一个为止

import java.util.*;
public class Solution {
    public int ysf (int n, int m) {
       List<Integer> list = new ArrayList<>();
       for(int i = 1 ; i<= n ;i++){
           list.add(i);
       }
       int begin = 0;
       while(list.size()!=1){
            begin = (m+begin-1)%list.size();
            list.remove(begin);
       }
       return list.get(0);
    }
}