zoukankan      html  css  js  c++  java
  • 趣味算法——青蛙过河(JAVA)

      青蛙过河是一个非常有趣的智力游戏,其大意如下: 一条河之间有若干个石块间隔,有两队青蛙在过河,每队有3只青蛙,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,不可以一次跳过两个位置,但是可以跳过对方一只青蛙进入到前面的一个空位。问两队青蛙该如何移动才能用最少的步数分别走向对岸?( → → → □ ← ← ← )可能3只青蛙太少了,心算也不难。如果有100只青蛙呢?

    /**
     * 青蛙过河
     * @author rubekid
     *
     */
    public class RiverFrog {
        
        public static final int LEFT_FROG = -1;
        
        public static final int RIGHT_FROG = 1;
        
        public static final int STONE = 0;
    
        private int[] frogs;
        
        private int zeroIndex;
        
        private int length;
    
        private int step = 0;
        
        public RiverFrog(int number) {
            frogs = new int[number * 2 +1];
            length = frogs.length;
            zeroIndex = length /2;
            for(int i=0; i< number; i++){
                frogs[i] = LEFT_FROG;
            }
            frogs[zeroIndex] = STONE;
            for(int i=0; i< number; i++){
                frogs[i+ zeroIndex + 1] = RIGHT_FROG;
            }
            
        }
        
        public void run(){
    
            while(!isMoveEnd(LEFT_FROG) || !isMoveEnd(RIGHT_FROG)){
                int left = zeroIndex - 1;
                int right = zeroIndex+1;
                
                if(left>-1 && right <length){
                    if(frogs[left] != frogs[right]){
                        if(frogs[left] == LEFT_FROG){
                            if(left > 0 && frogs[left-1] == RIGHT_FROG){//若移动right,则在中间有两只RIGHT并排
                                this.move(right);
                            }
                            else{
                                this.move(left);
                            }
                        }
                        else if(left > 0 && frogs[left-1]==LEFT_FROG ){
                            this.move(left-1);
                        }
                        else if(right <= length && frogs[right+1] == RIGHT_FROG){
                            this.move(right+1);
                        }
                    }
                    else{
                        if(frogs[left] == RIGHT_FROG){
                            if(left > 0 && frogs[left-1] == LEFT_FROG){
                                this.move(left - 1);
                            }
                            else if(right+1 < length && frogs[right+1] == RIGHT_FROG){
                                this.move(right+1);
                            }
                            else if(frogs[right] == RIGHT_FROG){
                                this.move(right);
                            }
                        }
                        else if(frogs[right] == LEFT_FROG){
                            if(right+1 < length && frogs[right+1] == RIGHT_FROG){
                                this.move(right + 1);
                            }
                            else if(left >0 && frogs[left-1] == LEFT_FROG){
                                this.move(left-1);
                            }
                            else if(frogs[left] == LEFT_FROG){
                                this.move(left);
                            }
                        }
                    }
                }
                else if(left == -1){
                    if(frogs[right] == LEFT_FROG && right<length-1){
                        this.move(right+1);
                    }
                    else{
                        this.move(right);
                    }
                }
                else if(right == length){
                    if(frogs[left] == RIGHT_FROG && left > 0){
                        this.move(left-1);
                    }
                    else{
                        this.move(left);
                    }
                }
            }
            System.out.println("step:" + step);
        }
    
        
        private void move(int i){
            int temp = frogs[i];
            frogs[i] = frogs[zeroIndex];
            frogs[zeroIndex] = temp;
            zeroIndex = i;
            step++;
            print();
        }
        
        private boolean isMoveEnd(int value){
            int i=0; int max= zeroIndex;
            if(value == LEFT_FROG){
                i = zeroIndex+1;
                max = length;
            }
            for(int j=i; j<max; j++){
                if(frogs[j]!=value){
                    return false;
                }
            }
            return true;
        }
        
        private void print(){
            StringBuffer stringBuffer = new StringBuffer();
            for(int frog : frogs){
                if(frog>-1){
                    stringBuffer.append(" " +frog + "    ");
                }
                else{
                    stringBuffer.append(frog + "    ");
                }
                
            }
            System.out.println(stringBuffer.toString());
        }
    }
  • 相关阅读:
    nodejs学习(一)--express+ejs生成项目
    react-native React Native version mismatch
    qrcode length overflow 生成二维码网址长度溢出解决办法
    禁止input输入空格
    微信开发:清除微信浏览器缓存
    多行文本溢出显示...的方法(-webkit-line-clamp)
    angular路由(自带路由篇)
    angular入门(基础篇)
    Python文件读写(open(),close(),with open() as f...)
    centos6.9使用yum安装mysql(简单粗暴,亲测有效)
  • 原文地址:https://www.cnblogs.com/rubekid/p/4207405.html
Copyright © 2011-2022 走看看