Java編程內(nèi)功-數(shù)據(jù)結(jié)構(gòu)與算法「環(huán)形鏈表與約瑟夫問(wèn)題」
Josephu問(wèn)題
設(shè)編號(hào)為1,2,....n的n個(gè)人圍坐一圈,約定編號(hào)為k(1<<k<<n)的人開(kāi)始報(bào)數(shù),數(shù)到m的那個(gè)人出列,它的下一位又從1開(kāi)始報(bào)數(shù),數(shù)到m的那個(gè)人又出列,依次類(lèi)推,知道所有人出列為止,由此產(chǎn)生一個(gè)出隊(duì)編號(hào)的序列.
循環(huán)鏈表處理Josephu問(wèn)題
先構(gòu)成一個(gè)有n個(gè)節(jié)點(diǎn)的單向循環(huán)鏈表,然后由k節(jié)點(diǎn)器從1開(kāi)始計(jì)數(shù),計(jì)到m時(shí),對(duì)應(yīng)節(jié)點(diǎn)從鏈表刪除,然后再?gòu)谋粍h除節(jié)點(diǎn)的下一個(gè)節(jié)點(diǎn)又從1開(kāi)始計(jì)數(shù),直到最后一個(gè)節(jié)點(diǎn)從鏈表中刪除.
構(gòu)建一個(gè)單向環(huán)形鏈表
1. 先創(chuàng)建第一個(gè)節(jié)點(diǎn),讓first指向該節(jié)點(diǎn),并形成環(huán).
2. 后面每創(chuàng)建一個(gè)新的節(jié)點(diǎn),就把該節(jié)點(diǎn),加入環(huán)形鏈表即可.
代碼案例
- package com.structures.linkedlist;
- public class Josephu {
- public static void main(String[] args) {
- CircleSingleLinkedList circleSingleLinkedList = new CircleSingleLinkedList();
- circleSingleLinkedList.addBoy(5);
- circleSingleLinkedList.showBoys();
- circleSingleLinkedList.countBoy(1,2,5);
- /*
- 小孩的編號(hào):1
- 小孩的編號(hào):2
- 小孩的編號(hào):3
- 小孩的編號(hào):4
- 小孩的編號(hào):5
- 小孩2出圈
- 小孩4出圈
- 小孩1出圈
- 小孩5出圈
- 最后留在圈中的小孩編號(hào)3
- */
- }
- }
- //創(chuàng)建一個(gè)環(huán)形的單向鏈表
- class CircleSingleLinkedList {
- //創(chuàng)建一個(gè)first節(jié)點(diǎn),當(dāng)前沒(méi)有編號(hào)
- private Boy first = new Boy(-1);
- //添加小孩節(jié)點(diǎn),構(gòu)建成一個(gè)環(huán)形鏈表
- public void addBoy(int nums) {
- if (nums < 1) {
- System.out.println("nums 值不正確");
- return;
- }
- Boy curBoy = null;
- //for循環(huán)創(chuàng)建環(huán)形鏈表
- for (int i = 1; i <= nums; i++) {
- Boy boy = new Boy(i);
- //如果是第一個(gè)小孩
- if (i == 1) {
- first = boy;
- first.setNext(first);
- curBoy = first;//讓curBoy指向第一個(gè)
- } else {
- curBoy.setNext(boy);
- boy.setNext(first);
- curBoy = boy;
- }
- }
- }
- //遍歷當(dāng)前環(huán)形鏈表
- public void showBoys() {
- if (first.getNext() == null) {
- System.out.println("沒(méi)有任何小孩~~");
- return;
- }
- Boy temp = first;
- while (true) {
- System.out.println("小孩的編號(hào):" + temp.getNo());
- if (temp.getNext() == first) {
- break;
- }
- temp = temp.getNext();
- }
- }
- /**
- * 根據(jù)用戶(hù)輸入,計(jì)算小孩出圈順序
- *
- * @param startNo 表示從第幾個(gè)小孩開(kāi)始計(jì)數(shù)
- * @param countNum 表示數(shù)幾下
- * @param nums 表示多少個(gè)小孩在圈中
- */
- public void countBoy(int startNo, int countNum, int nums) {
- //先進(jìn)行數(shù)據(jù)校驗(yàn)
- if (first == null || startNo < 1 || startNo > nums) {
- System.out.println("參數(shù)輸入有誤,請(qǐng)重新輸入");
- return;
- }
- //創(chuàng)建一個(gè)輔助指針,幫助完成小孩出圈
- Boy helper = first;
- //讓helper指向環(huán)形鏈表的最后節(jié)點(diǎn)
- while (helper.getNext() != first) {
- helper = helper.getNext();
- }
- //報(bào)數(shù)前,先讓helper和first移動(dòng),移動(dòng)k-1次,這樣first定位到開(kāi)始節(jié)點(diǎn),helper緊接著first
- for (int i = 0; i < startNo - 1; i++) {
- first = first.getNext();
- helper = helper.getNext();
- }
- //報(bào)數(shù)時(shí),讓first和helper指針同時(shí)移動(dòng),然后出圈
- while (true) {
- //當(dāng)圈中只有一個(gè)節(jié)點(diǎn)
- if (helper == first) {
- break;
- }
- //讓first和helper指針同時(shí)移動(dòng)countNum - 1次
- for (int i = 0; i < countNum - 1; i++) {
- first = first.getNext();
- helper = helper.getNext();
- }
- //此時(shí)first節(jié)點(diǎn)就是小孩要出圈的節(jié)點(diǎn)
- System.out.printf("小孩%d出圈\n", first.getNo());
- first = first.getNext();
- helper.setNext(first);
- }
- System.out.printf("最后留在圈中的小孩編號(hào)%d \n", first.getNo());
- }
- }
- //創(chuàng)建一個(gè)Boy類(lèi),表示節(jié)點(diǎn)
- class Boy {
- private int no;//編號(hào)
- private Boy next;//指向下一個(gè)節(jié)點(diǎn),默認(rèn)null
- public Boy(int no) {
- this.no = no;
- }
- public int getNo() {
- return no;
- }
- public void setNo(int no) {
- this.no = no;
- }
- public Boy getNext() {
- return next;
- }
- public void setNext(Boy next) {
- this.next = next;
- }
- }
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