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