小孩围圈数数+java,java实现n个小孩围圈数m退1有关问题（单链)

转载请注明出处：http://shuiguaiqq.iteye.com/blog/2065943

n个人围一圈报数，数到m的人退出，直到最后只剩一个人。

问题也就不详细描述了，百度一搜一大堆，以前看过马士兵的视频，里面讲到过用的好像类似双链，有left和right的，我自己也实现过，但是搞来搞去容易让看代码的看晕，数组实现也搞过，感觉也不够直观，我比较喜欢结构逻辑清晰的代码，所以感觉还是单链清爽点，代码如下：

/**

* n个人围一圈报数，数到m的人退出，直到最后只剩一个人

*/

public class CountQuit {

static final int N = 500;

static final int M = 3;

static int remainPeople;

static Node root = null;

static {

createCircle();

}

public static void main(String[] args) {

startGrame();

}

public static void startGrame() {

Node start = root;

int count = 1;

while (remainPeople > 1) {

if (count == CountQuit.M - 1) {

count = 0;

start.setNext(start.getNext().getNext());

remainPeople--;

}

start = start.getNext();

count++;

}

System.out.println("最后剩下的小孩编号为:"+start.getCode());

}

public static void createCircle() {

Node temp = null;

for (int i = 0; i < CountQuit.N; i++) {

if (i == 0) {

root = new Node(i + 1);

temp = root;

} else {

temp.setNext(new Node(i + 1));

temp = temp.getNext();

if (i == CountQuit.N - 1) {

temp.setNext(root);

}

}

}

remainPeople = CountQuit.N;

}

}

class Node {

private int code;

Node next;

public Node(int code) {

this.code = code;

}

public int getCode() {

return code;

}

public void setCode(int code) {

this.code = code;

}

public Node getNext() {

return next;

}

public void setNext(Node next) {

this.next = next;

}

}