演算法-經典趣題-青蛙過河
阿新 • • 發佈:2021-09-05
本文為joshua317原創文章,轉載請註明:轉載自joshua317部落格https://www.joshua317.com/article/90
一、問題
青蛙過河是一個非常有趣的智力遊戲,其大意如下:
一條河之間有若干石塊間隔,有兩隊青蛙在過河,每隊有3只青蛙,如圖所示。這些青蛙只能向前移動,不能向後移動,且一次只能有一隻青蛙向前移動。在移動過程中,青蛙可以向前面的空位中移動,不可一次跳過兩個位置,但是可以跳過對方一隻青蛙進入前面的一個空位。問兩隊青蛙該如何移動才能夠用最少的步數分別走向對岸?
二、分析
我們來分析一下青蛙過河問題。可以採用如下方案來移動青蛙,操作步驟如下:
(1)左側的青蛙向右跳過右側的一隻青蛙,落入空位,執行第(5)步。
(2)右側的青蛙向左跳過左側的一隻青蛙,落入空位,執行第(5)步。
(3)左側的青蛙向右移動一格,落入空位,執行第(5)步。
(4)右側的青蛙向左移動一格,落入空位,執行第(5)步。
(5)判斷是否已將兩隊青蛙移到對岸,如果沒有則繼續從第(1)步執行,否則結束程式。
三、程式設計
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("青蛙跳躍的順序為:\r\n " + frogJumpInfo); } } class Frog { static enum frogDirection {向左, 向右}; public String frogName;//青蛙名稱 public int position;//青蛙位置 public frogDirection direction;//青蛙跳動的方向 public boolean canJump;//是否可以跳 public boolean isEmpty = false;//是否是空格 //建構函式 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 { //初始化青蛙佇列 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; } //當一個青蛙跳動後,形成一個新的佇列 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; } //是否已經完成位置對換,即前三個青蛙的位置都大於3 private boolean isComplete(List<Frog> frogQueue) { return (frogQueue.get(0).position > 3 && frogQueue.get(1).position > 3 && frogQueue.get(2).position > 3); } //是否還有可以跳動的青蛙,只有可以跳動的,就沒有達到最後的狀態, //但都不可以跳動了也不一定對換完了,這裡只是控制遞迴 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; } //獲取青蛙跳動的步驟 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) + "個位置" + "\r\n"; int newPositionId = frog.position; List<Frog> newFrogQueue = this.editFrogQueue(frogQueue, frog.frogName, emptyPositionId,newPositionId); //只要能繼續跳就遞迴 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、練習
大家可以想一想如果是4只青蛙,5只青蛙,6只青蛙呢?
附一個小遊戲的連結:http://www.4399.com/flash/204168_4.htm
本文為joshua317原創文章,轉載請註明:轉載自joshua317部落格https://www.joshua317.com/article/90