合并两个数组

【题目】有两个排序的数组A1和A2，内存在A1的末尾有足够的空余空间容纳A2。请实现一个函数，把A2中所有数字插入到A1中，并且所有的数字是排序的。

1. 从后往前比较A1和A2中的数字，并把较大的数字复制到A1的合适位置。

public class Main {

    public static void main(String[] args) {

        int[] arr1 = new int[100];
        arr1[0] = 1;
        arr1[1] = 3;
        arr1[2] = 4;
        arr1[3] = 7;
        arr1[4] = 9;
        arr1[5] = 10;
        arr1[6] = 23;
        arr1[7] = 67;
        int[] arr2 = { 4, 6, 8, 10, 22, 24, 26, 34, 37 };

        Main main = new Main();
        int[] result = main.process(arr1, arr2);

        if (null != result) {
            for (int i = 0; i < (8 + arr2.length); i++) {
                System.out.print(result[i] + " ");
            }
        }
    }

    public int[] process(int[] arr1, int[] arr2) {

        if (null == arr1 || null == arr2
                || (0 == arr1.length && 0 == arr2.length)) {
            return null;
        }

        if (0 == arr1.length && 0 != arr2.length) {
            return arr2;
        }

        if (0 != arr1.length && 0 == arr2.length) {
            return arr1;
        }

        // 求arr1中非0的元素个数
        int index = 0;
        for (int i = 0; i < arr1.length; i++) {
            if (0 == arr1[i]) {
                index = i;
                break;
            }
        }

        int p1 = index - 1; // arr1末尾指针
        int p2 = arr2.length - 1; // arr2末尾指针
        int p = index + arr2.length - 1; // arr3末尾指针

        // 从后往前遍历arr1和arr2中的元素，并从后往前赋值到arr1中
        while (p1 >= 0 && p2 >= 0) {
            if (arr1[p1] >= arr2[p2]) {
                arr1[p--] = arr1[p1--];
            } else {
                arr1[p--] = arr2[p2--];
            }
        }

        while (p1 >= 0) {
            arr1[p--] = arr1[p1--];
        }

        while (p2 >= 0) {
            arr1[p--] = arr2[p2--];
        }

        return arr1;
    }
}