Question

前言

通过在我的日程安排中找到一些空闲时间，我不得不自我提高我的递归技能（不幸的是）。作为实践，我想通过使用递归来重新创建所有运算符，第一个是添加。虽然我有点卡住了。

问题

如上所述，我想通过仅使用递归和条件来重新创建加法运算符。虽然我完成了很多代码，但仍然存在一个问题，因为我包含了一个加法运算符。这是代码（运行正常，并在正，负和零输入的所有变体中按预期添加）。我还将一些平庸的评论作为帮助。

public class Test {
    public static void main(String[] args) {
        // Numbers to add
        int firstNumb = -5, secondNumb = 3;
        // Call the add function and save the result
        int result = add(firstNumb, secondNumb);
        // Print result
        System.out.println(result);
    }

    /*
     * Function recursively takes a number from 'giver' one at a time and
     * "gives"/"adds" it to 'receiver'. Once nothing more to "give" (second == 0),
     * then return the number that received the value, 'receiver'.
     */
    public static int add(int receiver, int giver) {
        /*
         * Base Case since nothing more to add on. != to handle signed numbers
         * instead of using > or <
         */
        if (giver != 0) {
            /*
             * Recursive Call.
             * 
             * The new 'giver' param is the incremental value of the number
             * towards 0. Ex: -5 -> -4 , 5 -> 4 (so I guess it may decrement).
             * 
             * The new 'receiver' param is the incremental value based on the
             * opposite direction the 'giver' incremented (as to why the
             * directionalIncrement() function needs both values to determine
             * direction.
             */
            return add(directionalIncrement(receiver, giver),
                    directionalIncrement(giver, -giver));
        } else {
            // Return 'receiver' which now contains all values from 'giver'
            return receiver;
        }
    }

    // Increments (or decrements) the 'number' based on the sign of the 'direction'
    public static int directionalIncrement(int number, int direction) {
        // Get incremental value (1 or -1) by dividing 'direction' by absolute
        // value of 'direction'
        int incrementalValue = direction / abs(direction);
        // Increment (or decrement I guess)
        return number + incrementalValue;
    }

    // Calculates absolute value of a number
    public static int abs(int number) {
        // If number is positive, return number, else make it positive by multiplying by -1 then return
        number = (number > 0.0F) ? number : -number;
        return number;
    }
}

问题是包含return number + incrementalValue;的行。如前所述，代码适用于此，但不符合我自己的规范，不涉及任何加法运算符。

我将行更改为return add(number, incrementalValue);，但似乎无法突破递归并确实抛出了此网站的标题，StackOverflowException.

所有帮助表示赞赏。提前谢谢。

注意

约束不包括任何隐式递增/递减（i ++ / i--），也不包括按位。尝试并回答我在自己的实现中遇到的具体问题。

Answer 1

public static int add(int a, int b) {
    if(b == 0) return a;
    int sum = a ^ b; //SUM of two integer is A XOR B
    int carry = (a & b) << 1;  //CARRY of two integer is A AND B
    return add(sum, carry);
}

羞辱地从here获取。所有功劳都归功于作者。

Answer 2

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

用++ / - 。

Java |仅使用递归和条件创建显式加法函数

2 个答案: