Java:如何对两个不同长度的数组的元素求和

Question

我想把两个不同长度的数组的元素加在一起。下面的代码仅适用于相同的长度，这是我目前所拥有的。

    //for the same lengths
int[]num1 = {1,9,9,9};
int[]num2 = {7,9,9,9};// {9,9,9}

    int total = 0, carry = 1;
    int capacity = Math.max(num1.length,num2.length);
    int []arraySum = new int [capacity];

        for (int i = capacity - 1 ; i >= 0; i--)
        {   
            arraySum[i] = num1[i]+ num2[i];

            if (arraySum[i] > 9)
            {
                arraySum[i] = arraySum[i] % 10;

                num2[i-1] = num2[i-1] + carry;  
            }


        }

    for(int i = 0; i < arraySum.length; i++)
    {
        System.out.print(arraySum[i]);
    }

把num2和length中的元素改成像{9,9,9}怎么办？我知道我可能需要将另一个 for 循环作为内部 for 循环并控制长度较小的数组的索引但是如何......？还有一件事......我应该为那些 for 循环条件做些什么，因为 num1 和 num2 最终将由用户输入。

好吧，你可以看出输入是有限的，因为如果 num1[0] + num2[0] > 9 进位没有要放置的索引，则无法编译。因此，我需要将整个数组向右移动并将进位从 num1[0] + num2[0] 放置。问题来了!!我应该把转移码放在哪里？我有点糊涂了......

Answer 1

实际上，您声明了一个容量为 Math.max(num1.length, num2.length) 的 int[] 数组。

还不够。您应该将容量设置为 Math.max(num1.length, num2.length) +1。

为什么？

看num1是否为{1,9,9,9}，num2是否为{9,9,9,9}，arraySum如何表示总和 {1,1,9,9,8}？

所以我们需要声明如下，考虑是否需要carry。

 int[] arraySum = new int[capacity + 1];

然后在打印sum的时候，检查arraySum[0]是0还是1，如果等于0，就不要在Console打印。

引用修改代码如下:

封装问题；

public class Example {

public static void main(String[] args) {

    // for the same lengths
    int[] num1 = { 1,9,9,9 };
    int[] num2 = { 9,9,9,9};// {9,9,9}

    // 1999+9999 = 11998, its length is greater than the max
    int capacity = Math.max(num1.length, num2.length);
    int[] arraySum = new int[capacity + 1];

    int len2 = num2.length;
    int len1 = num1.length;

    if (len1 < len2) {
        int lengthDiff = len2 - len1;

        /*
         * Flag for checking if carry is needed.
         */
        boolean needCarry = false;

        for (int i = len1 - 1; i >= 0; i--) {
            /**
             * Start with the biggest index
             */
            int sumPerPosition =0;

            if (needCarry) {
                 sumPerPosition = num1[i] + num2[i + lengthDiff] +1;
                 needCarry = false;
            }else
            {
                 sumPerPosition = num1[i] + num2[i + lengthDiff];
            }

            if (sumPerPosition > 9) {
                    arraySum[i + lengthDiff + 1] = sumPerPosition % 10;
                needCarry = true;
            }else
            {
                arraySum[i + lengthDiff + 1] = sumPerPosition % 10;
            }
        }

        /**
         * Handle the remaining part in nun2 Array
         */

        for (int i = lengthDiff - 1; i >= 0; i--) {
            /*
             * Do not need to care num1 Array Here now
             */

            if(needCarry){
                arraySum[i + 1] = num2[i]+1;
            }else
            {
                arraySum[i + 1] = num1[i] ;
            }

            if (arraySum[i + 1] > 9) {
                arraySum[i + 1] = arraySum[i + 1] % 10;
                needCarry = true;
            } else {
                needCarry = false;
            }


        }

        /*
         * Handle the last number, if carry is needed. set it to 1, else set
         * it to 0
         */
        if (needCarry) {
            arraySum[0] = 1;
        } else {
            arraySum[0] = 0;
        }

    } else {
        int lengthDiff = len1 - len2;

        /*
         * Flag for checking if carry is needed.
         */
        boolean needCarry = false;

        for (int i = len2 - 1; i >= 0; i--) {
            /**
             * Start with the biggest index
             */
            int sumPerPosition = 0;


            if (needCarry) {
                 sumPerPosition = num2[i] + num1[i + lengthDiff] +1;
                 needCarry = false;
            }else
            {
                 sumPerPosition = num2[i] + num1[i + lengthDiff];
            }

            if (sumPerPosition > 9) {
                    arraySum[i + lengthDiff + 1] = sumPerPosition % 10;
                needCarry = true;
            }else
            {
                arraySum[i + lengthDiff + 1] = sumPerPosition % 10;
            }
        }

        /**
         * Handle the remaining part in nun2 Array
         */

        for (int i = lengthDiff - 1; i >= 0; i--) {
            /*
             * Do not need to care num1 Array Here now
             */

            if(needCarry){
                arraySum[i + 1] = num1[i]+1;
            }else
            {
                arraySum[i + 1] = num1[i] ;
            }

            if (arraySum[i + 1] > 9) {
                arraySum[i + 1] = arraySum[i + 1] % 10;
                needCarry = true;
            } else {
                needCarry = false;
            }

        }

        /*
         * Handle the last number, if carry is needed. set it to 1, else set
         * it to 0
         */
        if (needCarry) {
            arraySum[0] = 1;
        } else {
            arraySum[0] = 0;
        }
    }

    /*
     * Print sum 
     * 
     * if arraySum[0] ==1, print 1
     * 
     * Do not print 0 when arraySum[0] ==0
     */
    if(arraySum[0] == 1)
    {
        System.out.print(1);
    }
    for (int i = 1; i < arraySum.length; i++) {

        System.out.print(arraySum[i]);
    }
}
}

以num1为{1,9,9,9}，num2为{9,9,9,9}为例，求和结果如下:

控制台输出:

11998