看题目提示使用链表，就顺便复习一下单向循环链表了

画个图更好理解些，偷个懒，不画了呢

class Node{
    int val;
    Node next = null;
    public Node(int val) {
        this.val = val;
    }
}

public class Solution {
    public int LastRemaining_Solution(int n, int m) {
        if(n<=0||m<=0)return -1;
        //1.将数字串起来，围成一个圈
        Node head = new Node(0);
        Node headP1 = head;
        for(int i=1;i<n;i++){
           headP1.next = new Node(i);
           headP1 = headP1.next;
        }
        headP1.next = head;
        //定义一个新的头节点开始数数
        Node headP2 = head;
        Node pre = null ;//前驱节点
        int count = 1;//记录当前走过的节点个数
        while(true){
            if(count==m){//当走过的个数为m个时开始删除当前节点，然后重新循环
                if(headP2.next == pre) return pre.val;//结束条件是：链表中只剩下两个节点
                //下面是删除节点过程
                Node temp = headP2.next;
                pre.next = temp;
                headP2.next = null;
                headP2 = temp;
                //从头重新计数
                count=1;
            }else{
                pre = headP2;
                headP2 = headP2.next;
                count++;
            }

        }


    }
}