淘宝面试题 n个人围成一圈，报到m的人出列

public class ReportNumber {

	public static void main(String[] args) throws Exception {
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		System.out.print("please input two numbers : ");
		
		String[] strings = br.readLine().replaceAll("\\s+", "").split(",");
		
		int m = Integer.parseInt(strings[0]);
		
		int n = Integer.parseInt(strings[1]);
		
		LinkedList<Integer> linkedList = new LinkedList<Integer>();
		
		for (int i = 1; i <= n; i++) {
			
			linkedList.add(i);
		}
		
		for (int i = 0, j = 1; !linkedList.isEmpty(); i++, j++) {
			
			// 如[报数报程中]下标指针到达链表尾部，须从头开始重新报数
			if (i == linkedList.size()) {
				
				i = 0;
			}
			
			if (j == m) {
				
				System.out.printf("%d ", linkedList.remove(i));
				
				// 如[报数完成后]下标指针下标指针到达链表尾部，须从头开始重新报数
				if (i == linkedList.size()) {
					
					i = 0;
				}
				
				j = 1;
			}
		}
	}
}

public int find(int n, int m)
	{
		int r = 0;
		for (int i = 2; i <= n; i++) {
			r = (r + m) % i;
		}
		return r + 1;
	}

public class Josephus {
	static class Node {
		int val;
		Node next;

		Node(int v) {
			val = v;
		}
	}// 成员类，代表节点，类似于数据结构中的结构体

	public static void main(String[] args) {
		int N = 9;// 这个表示总人数
		int M = 5;// 数到几的人出列
		Node t = new Node(1);// 头节点单列出来，方便形成循环链表
		Node x = t;

		for (int i = 2; i <= N; i++)
			x = (x.next = new Node(i));// 建立单向链表
		x.next = t;// 最后一个节点的next指向第一个节点，形成循环链表
		System.out.println("出圈的顺序为：");
		while (x != x.next) {
			for (int i = 1; i < M; i++)
				x = x.next;
			// 此时x是将出列的节点的前一个节点
			System.out.print(x.next.val + " ");
			x.next = x.next.next;
		}
		System.out.println();
		System.out.println("Survivors is " + x.val);
	}// end main
}