约瑟夫环问题

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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;    }}

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