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 https://www.joshua317.com/article/90

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：
 " + 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3
    private 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" +
                    "
";
            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 https://www.joshua317.com/article/90