java 解决倒水问题

问题导入：

倒水问题描述：现在有3L，4L的杯子，无限多的水，要求用这两个容器倒出5L的水，如何求解?

求解具体代码：

/**
 * Question:倒水问题描述：现在有3L，4L的杯子，无限多的水，要求用这两个容器倒出5L的水，如何求解?
 * Author: ChenPeng
 * CompleteTime:2019-04-21
 */

import java.util.*;


/**
 * 状态的数据结构(a,b)
 * parentNode 用于最终路径回溯
 */
class WaterStatus implements Cloneable {

    private int a; // a容器
    private int b; // b容器

    private WaterStatus parentNode; // 父节点

    public WaterStatus(int a, int b) {
        this.a = a;
        this.b = b;
        this.parentNode = null;
    }

    public void setA(int a) {
        this.a = a;
    }

    public int getA() {
        return this.a;
    }

    public void setB(int b) {
        this.b = b;
    }

    public int getB() {
        return this.b;
    }

    public WaterStatus getParentNode() {
        return parentNode;
    }

    public void setParentNode(WaterStatus parentNode) {
        this.parentNode = parentNode;
    }

    /**
     * 判断是否结束
     *
     * @return
     */
    public boolean isEnd() {
        return this.a + this.b == 5;
    }

    /**
     * 重写equals方法
     *
     * @param obj
     * @return
     */
    @Override
    public boolean equals(Object obj) {
        if (!(obj instanceof WaterStatus)) {
            return false;
        }
        if (this == obj) {
            return true;
        }
        boolean flag = (((WaterStatus) obj).getA() == this.getA())
                && (((WaterStatus) obj).getB() == this.getB());
        return flag;
    }

    /**
     * 重写hashcode方法
     *
     * @return
     */
    @Override
    public int hashCode() {
        int result = Integer.hashCode(a);
        result = 31 * result + Integer.hashCode(b);
        return result;
    }

    /**
     * 克隆方法（浅拷贝）
     *
     * @return
     */
    @Override
    protected Object clone() {
        WaterStatus waterStatus = null;
        try {
            waterStatus = (WaterStatus) super.clone();
        } catch (CloneNotSupportedException e) {
            e.printStackTrace();
        }
        return waterStatus;
    }

    /**
     * 重写toString方法
     *
     * @return
     */
    @Override
    public String toString() {
        return "(" + this.a + "," + this.b + ")";
    }
}

/**
 * 倒水问题求解
 */
public class DaoShui {

    // A容器的容量
    private static final int A_CONTAINER = 3;

    // B容器的容量
    private static final int B_CONTAINER = 4;

    // 历史状态存储
    private static Set histroyStatus = new HashSet();

    public static void main(String[] args) {
        // 初始状态
        WaterStatus water = new WaterStatus(0, 0);
        histroyStatus.add(water);
        // 调用求解5L的路径
        calculatePath(water);
    }


    /**
     * 计算路径（核心方法）
     *
     * @param water
     */
    public static void calculatePath(WaterStatus water) {
        if (water.isEnd()) { // 算法终止条件是满足最终需要倒出5L水
            // 最终路径序列(栈)先进后出
            Stack resultPath = new Stack<>();
            while (water != null) {
                resultPath.push(water); // 压入栈中
                water = water.getParentNode();
            }
            // 输出结果
            printStack(resultPath);
            return;
        }

        // 存储每一层的中间结果
        List tempList = new ArrayList<>();


        // a <-
        WaterStatus waterStatus1 = (WaterStatus) water.clone();
        // 这里判断是否能采用该方法&&该方法后不能与之前的状态重合(下同)
        if (aGetWater(waterStatus1) && (!histroyStatus.contains(waterStatus1))) {
            histroyStatus.add(waterStatus1); // 添加历史记录(下同)
            waterStatus1.setParentNode(water); // 设置父节点
            tempList.add(waterStatus1); // 保存中间层结果(下同)
        } else {
            waterStatus1 = null;
        }

        // b <-
        WaterStatus waterStatus2 = (WaterStatus) water.clone();
        if (bGetWater(waterStatus2) && (!histroyStatus.contains(waterStatus2))) {
            histroyStatus.add(waterStatus2);
            waterStatus2.setParentNode(water);
            tempList.add(waterStatus2);
        } else {
            waterStatus2 = null;
        }

        // a -> b
        WaterStatus waterStatus3 = (WaterStatus) water.clone();
        if (aTobWater(waterStatus3) && (!histroyStatus.contains(waterStatus3))) {
            histroyStatus.add(waterStatus3);
            waterStatus3.setParentNode(water);
            tempList.add(waterStatus3);
        } else {
            waterStatus3 = null;
        }

        // b -> a
        WaterStatus waterStatus4 = (WaterStatus) water.clone();
        if (bToaWater(waterStatus4) && (!histroyStatus.contains(waterStatus4))) {
            histroyStatus.add(waterStatus4);
            waterStatus4.setParentNode(water);
            tempList.add(waterStatus4);
        } else {
            waterStatus4 = null;
        }


        // a ->
        WaterStatus waterStatus5 = (WaterStatus) water.clone();
        if (aEmpty(waterStatus5) && (!histroyStatus.contains(waterStatus5))) {
            histroyStatus.add(waterStatus5);
            waterStatus5.setParentNode(water);
            tempList.add(waterStatus5);
        } else {
            waterStatus5 = null;
        }


        // b ->
        WaterStatus waterStatus6 = (WaterStatus) water.clone();
        if (bEmpty(waterStatus6) && (!histroyStatus.contains(waterStatus6))) {
            histroyStatus.add(waterStatus6);
            waterStatus6.setParentNode(water);
            tempList.add(waterStatus6);
        } else {
            waterStatus6 = null;
        }

        // 强制通知开始垃圾回收,但是不会立即执行
        System.gc();

        if (tempList.size() > 0) {
            // 然后依次递归调用
            for (WaterStatus e : tempList) {
                calculatePath(e);
            }
        }
    }


    /**
     * a取水              a <-
     *
     * @param water
     * @return
     */
    public static boolean aGetWater(WaterStatus water) {
        if (water.getA() == A_CONTAINER) {
            return false;
        }
        water.setA(A_CONTAINER);
        return true;
    }

    /**
     * b取水              b <-
     *
     * @param water
     * @return
     */
    public static boolean bGetWater(WaterStatus water) {
        if (water.getB() == B_CONTAINER) {
            return false;
        }
        water.setB(B_CONTAINER);
        return true;
    }

    /**
     * a把水倒入b中        a -> b
     *
     * @param water
     * @return
     */
    public static boolean aTobWater(WaterStatus water) {
        int a = water.getA();
        int b = water.getB();
        if (a == 0 || b == B_CONTAINER) {
            return false;
        }
        if (a + b > B_CONTAINER) {
            water.setB(B_CONTAINER);
            water.setA(a + b - B_CONTAINER);
        } else {
            water.setA(0);
            water.setB(a + b);
        }
        return true;
    }

    /**
     * b把水倒入a中       b -> a
     *
     * @param water
     * @return
     */
    public static boolean bToaWater(WaterStatus water) {
        int a = water.getA();
        int b = water.getB();
        if (b == 0 || a == A_CONTAINER) {
            return false;
        }
        if (a + b > A_CONTAINER) {
            water.setA(A_CONTAINER);
            water.setB(a + b - A_CONTAINER);
        } else {
            water.setB(0);
            water.setB(a + b);
        }
        return true;
    }

    /**
     * 倒出a中的水        a ->
     *
     * @param water
     * @return
     */
    public static boolean aEmpty(WaterStatus water) {
        if (water.getA() == 0) {
            return false;
        }
        water.setA(0);
        return true;
    }

    /**
     * 倒出b中的水        b ->
     *
     * @param water
     * @return
     */
    public static boolean bEmpty(WaterStatus water) {
        if (water.getB() == 0) {
            return false;
        }
        water.setB(0);
        return true;
    }


    /**
     * 输出最终结果
     *
     * @param stacks
     */
    public static void printStack(Stack stacks) {
        System.out.println("\n============倒水问题的路径为============");
        if (stacks != null) {
            while (!stacks.empty()) {
                System.out.print(stacks.pop() + "->");
            }
            System.out.println("end.");
        }
        System.out.println("=================结束=================\n");

    }

}

涉及知识点：

java深拷贝与浅拷贝；

Stack、HashSet集合的使用；

重写HashCode和equals方法；

 

运行结果：