Question

我有x = 10和y = 6的值。然后，我有这个addxy函数来计算两个数字的总和。我在这个添加中尽我所能，现在我正在研究如何构建它以获得产品.... 我已经把我的大脑震撼了一天，但我仍然无法得到它......

public static int addxy(int a, int b)
{
    if (b==0)
        return a;
    else
        if (b>0)
            return 1 + addxy(a,b-1);
        else
            if (a==0)
                return b;
            else
                return 1 + addxy(a-1,b);
}

任何帮助将不胜感激

Answer 1

我想你想要这样的东西，对吧？

public static int multiply(int a, int b)
{
    if (a == 0 || b == 0)
        return 0;
    if (a < 0)
        return -multiply(-a, b);
    if (b < 0)
        return -multiply(a, -b);

    return multiply(a, b - 1) + a;
}

Answer 2

如果您想拥有x*y，那么您必须递归添加x，y times，如下所示：

public class Sample {
public static void main(String[] args) {
    System.out.println(add(5, 6));
    System.out.println(add(-5, 6));
    System.out.println(add(5, -6));
    System.out.println(add(-5, -6));
}

static int add(int x, int y) {

    if (x==0 ||y == 0) {
        return 0;
    }
    if (y > 0)
        return x + add(x, y - 1); //for positive numbers 
    else
        return -x + add(x, y + 1); // for negative numbers

}
}

O / P：

Answer 3

public static int mulxy( int a, int b ) {
    if ( b == 0 || a == 0 ) {
        return 0;
    }

    // We need to solve the sign problem.         
    int signA = ( a > 0) ? 1 : -1;
    int signB = ( b > 0 )? 1 : -1;

    // now we know the sign of result.
    int signResult = signA * signB;
    // we don't need sign anymore
    int posA = a * signA;
    int posB = b * signB;

    return signResult * mulPosAB( posA, posB );   
}

public int multPosAB( int a, int b ) {
    // here we already know that a and b not -tive.

    if( a == 0 || b == 0 ) return 0;

    // return a + mulPosAB( a, b - 1 );
    // or using newAddXY
    return newAddXY( a, mulPosAB( a, b - 1 ) );

}

对于a和b均为负数的情况，您的addxy也不会工作。

public int newAddXY( int a, int b ) {
    if( b == 0 ) return a;
    if( a == 0 ) return b;

    // We need to solve the sign problem.         
    int signA = ( a > 0) ? 1 : -1;
    int signB = ( b > 0 )? 1 : -1;

    //return signB + newAddXY( a, b - signB );  
    // I think we should ensure we re not adding numbers bigger than 1.
    // so this will be better
    return newAddXY( signB, newADDXY( a, b-signB ) );
}

在乘法中使用递归

3 个答案: