算法-经典趣题-青蛙过河

本文为joshua317原创文章,转载请注明：转载自joshua317博客算法-经典趣题-青蛙过河 - joshua317的博客

一、问题

青蛙过河是一个非常有趣的智力游戏，其大意如下：

一条河之间有若干石块间隔，有两队青蛙在过河，每队有3只青蛙，如图所示。这些青蛙只能向前移动，不能向后移动，且一次只能有一只青蛙向前移动。在移动过程中，青蛙可以向前面的空位中移动，不可一次跳过两个位置，但是可以跳过对方一只青蛙进入前面的一个空位。问两队青蛙该如何移动才能够用最少的步数分别走向对岸？

二、分析

我们来分析一下青蛙过河问题。可以采用如下方案来移动青蛙，操作步骤如下：

（1）左侧的青蛙向右跳过右侧的一只青蛙，落入空位，执行第（5）步。

（2）右侧的青蛙向左跳过左侧的一只青蛙，落入空位，执行第（5）步。

（3）左侧的青蛙向右移动一格，落入空位，执行第（5）步。

（4）右侧的青蛙向左移动一格，落入空位，执行第（5）步。

（5）判断是否已将两队青蛙移到对岸，如果没有则继续从第（1）步执行，否则结束程序。

三、编程

package com.joshua317;import jdk.nashorn.internal.ir.LiteralNode;import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;public class Main {public static void main(String[] args) {FrogCrossRiver frogCrossRiver = new FrogCrossRiver();List frogQueue = frogCrossRiver.initFrogQueue();String frogJumpInfo = (frogCrossRiver.frogJump(frogQueue, 3));System.out.println("青蛙跳跃的顺序为：\r\n " + frogJumpInfo);}
}class Frog {static enum frogDirection {向左, 向右};public String frogName;//青蛙名称public int position;//青蛙位置public frogDirection direction;//青蛙跳动的方向public boolean canJump;//是否可以跳public boolean isEmpty = false;//是否是空格//构造函数public Frog (int position, String frogName, frogDirection direction, boolean canJump) {this.position = position;this.frogName = frogName;this.direction = direction;this.canJump = canJump;}public Frog (int position) {this.frogName = "空";this.position = position;this.canJump = false;this.isEmpty = true;}public Frog (Frog frog) {this.position = frog.position;this.frogName = frog.frogName;this.direction = frog.direction;this.canJump = frog.canJump;this.isEmpty = frog.isEmpty;}}class FrogCrossRiver {//初始化青蛙队列public List<Frog> initFrogQueue(){List<Frog> frogQueue = new ArrayList<Frog>();frogQueue.add(new Frog(0, "左1", Frog.frogDirection.向右, false));frogQueue.add(new Frog(1, "左2", Frog.frogDirection.向右, true));frogQueue.add(new Frog(2, "左3", Frog.frogDirection.向右, true));frogQueue.add(new Frog(3));frogQueue.add(new Frog(4, "右1", Frog.frogDirection.向左, true));frogQueue.add(new Frog(5, "右2", Frog.frogDirection.向左, true));frogQueue.add(new Frog(6, "右3", Frog.frogDirection.向左, false));return frogQueue;}//当一个青蛙跳动后，形成一个新的队列private List<Frog> editFrogQueue(List<Frog> frogQueue, String frogName, int oldEmptyPostionId, int newEmptyPostionId){List<Frog> newFrogQueue = new ArrayList<Frog>();for (int i=0; i<frogQueue.size(); i++) {Frog frog = (Frog)frogQueue.get(i);Frog newFrog = new Frog(frog);if (newFrog.isEmpty) {newFrog.position = newEmptyPostionId;}if (newFrog.frogName == frogName) {newFrog.position = oldEmptyPostionId;}newFrog.canJump = false;if ((newEmptyPostionId - newFrog.position) > 0 &&(newEmptyPostionId - newFrog.position) < 3 &&newFrog.direction == Frog.frogDirection.向右) {newFrog.canJump = true;}if ((newFrog.position - newEmptyPostionId) > 0 &&(newFrog.position - newEmptyPostionId) < 3 &&newFrog.direction == Frog.frogDirection.向左) {newFrog.canJump = true;}newFrogQueue.add(newFrog);}return newFrogQueue;}//是否已经完成位置对换，即前三个青蛙的位置都大于3private boolean isComplete(List<Frog> frogQueue){return (frogQueue.get(0).position > 3 && frogQueue.get(1).position > 3 && frogQueue.get(2).position > 3);}//是否还有可以跳动的青蛙，只有可以跳动的，就没有达到最后的状态，//但都不可以跳动了也不一定对换完了，这里只是控制递归private boolean canFrogJump(List<Frog> frogQueue){for (int i=0; i<frogQueue.size(); i++) {Frog frog = (Frog)frogQueue.get(i);if (frog.canJump) {return true;}}return false;}//获取青蛙跳动的步骤public String frogJump(List<Frog> frogQueue, int emptyPositionId){String frogJumpInfo = "";for (int i=0; i<frogQueue.size(); i++) {Frog frog = (Frog)frogQueue.get(i);//是空位置if (frog.isEmpty) {continue;}//不能跳if (!frog.canJump) {continue;}frogJumpInfo = "青蛙" + frog.frogName + " " + frog.direction + "跳到第" + (emptyPositionId + 1) + "个位置" +"\r\n";int newPositionId = frog.position;List<Frog> newFrogQueue = this.editFrogQueue(frogQueue, frog.frogName, emptyPositionId,newPositionId);//只要能继续跳就递归if (this.canFrogJump(newFrogQueue)) {frogJumpInfo += this.frogJump(newFrogQueue, newPositionId);} else {if (this.isComplete(newFrogQueue)) {frogJumpInfo += "成功";break;}}if (frogJumpInfo.contains("成功")) {break;}}return frogJumpInfo;}
}

四、练习

大家可以想一想如果是4只青蛙，5只青蛙，6只青蛙呢？

附一个小游戏的链接：http://www.4399.com/flash/204168_4.htm

本文为joshua317原创文章,转载请注明：转载自joshua317博客算法-经典趣题-青蛙过河 - joshua317的博客