OJ-数组形式的整数加法

题目含义举例：

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* addToArrayForm(int* num, int numSize, int k, int* returnSize) {
    int* res = malloc(sizeof(int) * fmax(10, numSize + 1));
    *returnSize = 0;
    for (int i = numSize - 1; i >= 0; --i) {
        int sum = num[i] + k % 10;
        k /= 10;
        if (sum >= 10) {
            k++; //进位
            sum -= 10; //进位后剩余的数值
        }
        res[(*returnSize)++] = sum;//逆序存储数据
    }

    //数组处理完后，k值还有剩余，直接将k取余逆序存入数组当中
    for (; k > 0; k /= 10) {
        res[(*returnSize)++] = k % 10;
    }
    for (int i = 0; i < (*returnSize) / 2; i++) {
        int tmp = res[i];
        res[i] = res[(*returnSize) - 1 - i];
        res[(*returnSize) - 1 - i] = tmp;
    }
    return res;
}