Question

从if和else.Calee递归调用相同的方法获取意外的输出（从sysout中可以看到）。预期产出：4

调试代码并看到发生了什么。但是不能理解。请帮忙

public class GetSquares {

    /**
     * @param args
     */
    public static void main(String[] args) {
        String rectDimension="5X3";
        int noOfSquares=0;
        int noOfSq=getNoofSquares(rectDimension,noOfSquares);
        System.out.println(noOfSq);
    }

    public static  int getNoofSquares(String rectDimension,int noOfSquares){

        int length=Integer.parseInt(rectDimension.split("X")[0]);
        int breadth=Integer.parseInt(rectDimension.split("X")[1]);


        if(length<breadth){
            rectDimension=String.valueOf(breadth-length)+"X"+length;
            ++noOfSquares;
            getNoofSquares(rectDimension,  noOfSquares);
        }else if(length>breadth){
            rectDimension=String.valueOf(length-breadth)+"X"+breadth;
            ++noOfSquares;
            getNoofSquares(rectDimension,  noOfSquares);
        }else  {
            noOfSquares=noOfSquares+1;
            System.out.println(noOfSquares);
            return noOfSquares;
        }

        return noOfSquares;



    }
}

Answer 1

你没有在递归调用中保存方法getNoofSquare()的返回值，得到像

这样的价值

noOfSquares= getNoofSquares(rectDimension,  noOfSquares);

Answer 2

您的问题是Java中的原始类型是按值传递的，而不是通过引用传递的。

public static void increment(int x) {
    x++;
}

public static void main(String... a) {
    int x = 0;
    increment(x);
    System.out.println(x);
}

此代码将打印0。

如果你想要一个“输出参数”，它必须是可变的，但int不是。您必须创建一个类，该类是整数的持有者：

public class MutableInteger {

    private int value = 0;

    public void increment() {
        value++;
    }

    public int getValue() {
        return value;
    }

}

然后你的代码看起来像这样：

// SOLUTION 1
public class GetSquares {

    public static void main(String[] args) {
        String rectDimension = "5X3";
        MutableInteger noOfSq = new MutableInteger();
        getNoofSquares(rectDimension, noOfSq);
        System.out.println(noOfSq.getValue());
    }

    public static void getNoofSquares(String rectDimension, MutableInteger noOfSquares) {

        int length = Integer.parseInt(rectDimension.split("X")[0]);
        int breadth = Integer.parseInt(rectDimension.split("X")[1]);

        noOfSquares.increment();

        if (length < breadth) {
            rectDimension = String.valueOf(breadth - length) + "X" + length;
            getNoofSquares(rectDimension, noOfSquares);
        } else if (length > breadth) {
            rectDimension = String.valueOf(length - breadth) + "X" + breadth;
            getNoofSquares(rectDimension, noOfSquares);
        }
    }

}

另一种方法是只使用返回值来返回结果：

// SOLUTION 2
public class GetSquares {

    public static void main(String[] args) {
        String rectDimension = "5X3";
        int noOfSq = getNoofSquares(rectDimension);
        System.out.println(noOfSq);
    }

    public static int getNoofSquares(String rectDimension) {
        int length = Integer.parseInt(rectDimension.split("X")[0]);
        int breadth = Integer.parseInt(rectDimension.split("X")[1]);

        if (length < breadth) {
            rectDimension = String.valueOf(breadth - length) + "X" + length;
            return 1 + getNoofSquares(rectDimension);
        } else if (length > breadth) {
            rectDimension = String.valueOf(length - breadth) + "X" + breadth;
            return 1 + getNoofSquares(rectDimension);
        } else {
            return 1;
        }
    }

}

此外，我建议只使用两个int长度和宽度参数，而不是进行字符串解析/拆分/构建：

// SOLUTION 2 (cleaned)
public class GetSquares {

    public static void main(String[] args) {
        String rectDimension = "5X3";

        int length = Integer.parseInt(rectDimension.split("X")[0]);
        int breadth = Integer.parseInt(rectDimension.split("X")[1]);

        int noOfSq = getNoofSquares(length, breadth);
        System.out.println(noOfSq);
    }

    public static int getNoofSquares(int length, int breadth) {
        if (length < breadth) {
            return 1 + getNoofSquares(breadth - length, length);
        } else if (length > breadth) {
            return 1 + getNoofSquares(length - breadth, breadth);
        } else {
            return 1;
        }
    }

}

Answer 3

Java是按值传递的。您正在递增一个局部变量，并且您没有对递归返回的值执行任何操作。

既然你没有做任何事情，但回来后你也可以做得很好：

public class GetSquares {

    public static void main(String[] args) {
        String rectDimension = args.length > 0 ? args[0] : "5X3";
        System.out.println("Number of squares from " + rectDimension 
            + " is " + getNoofSquares(rectDimension));
    }

    public static  int getNoofSquares(String rectDimension){
        String parts[] = rectDimension.split("X");
        int length=Integer.parseInt(parts[0]);
        int breadth=Integer.parseInt(parts[1]);
        return getNoofSquares(length, breadth, 0);
    }

    public static  int getNoofSquares(int length, int breadth, int noOfSquares){
        if (length < breadth) {
            return getNoofSquares(breadth-length, length, noOfSquares+1);
        } else if (length > breadth) {
            return getNoofSquares(length-breadth, breadth, noOfSquares+1);
        } else {
            return noOfSquares+1;
        }
    }
}

我保留了你的累加器所以现在它的尾部递归。但Java还没有TCO。

从if和else.Getting意外输出递归调用相同的方法

3 个答案: