约瑟夫环问题

n个人编号为1,2…n,每次数到m,编号为m的人就退出，问最后留下的人的编号是多少

package com.lxm.algorithm;
import java.util.Scanner;
public class YSF
{
    public static void main(String[] args)
    {
            Scanner scanner = new Scanner(System.in);
            while(scanner.hasNext())
            {
                int n = scanner.nextInt();
                int m = scanner.nextInt();
                int result = ysf(n,m);
                System.out.println(result);
            }


    }

    public static int  ysf(int n, int m)
    {
        //initialize array
        int[] data  = new int[n];
        for(int i=0;i<n;i++)
        {
            data[i] = i+1;
        }


        int left = n;//lasted people
        int index = -1;
        int num=0;

        while(left>1)
        {
                index =( index+1)%n;
                if(data[index]!=0)
                {
                    ++num;
                    if(num==m)
                    {
                        data[index] = 0;
                        --left;
                        num=0;
                    }
                }
        }
        for(int i=0;i<n;i++)
        {
            if(data[i]!=0)
            {
                return i+1;
            }
        }

        return -1;

    }
}