请帮我写一段能实现一下要求的c++语言代码：将两个非递减的有序链表合并为一个非递增的有序链表。要求结果链表仍使用原来两个链表的存储空间，不另外占用其他的存储空间。表中允许有重复的数据。数据内容自定义，类型为整型数据，每组十个数据，需提供2各测试用例，即共四组数据

以下是实现题目要求的C语言代码：

#include <stdio.h>

typedef struct node {
    int value;
    struct node *next;
} Node;

void merge(Node *list1, Node *list2) {
    Node *p1 = list1->next;
    Node *p2 = list2->next;
    Node *temp = list1;

    while (p1 != NULL && p2 != NULL) {
        if (p1->value >= p2->value) {
            temp->next = p1;
            p1 = p1->next;
        } else {
            temp->next = p2;
            p2 = p2->next;
        }
        temp = temp->next;
    }

    if (p1 != NULL) {
        temp->next = p1;
    } else {
        temp->next = p2;
    }

    Node *prev = NULL;
    Node *curr = list1->next;
    Node *next;

    while (curr != NULL) {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }

    list1->next = prev;
}

void printList(Node *list) {
    Node *p = list->next;
    while (p != NULL) {
        printf("%d ", p->value);
        p = p->next;
    }
    printf("\n");
}

int main() {
    // 第一组测试数据
    Node *list1_1 = (Node *) malloc(sizeof(Node));
    list1_1->next = NULL;
    Node *p = list1_1;
    int i;
    for (i = 1; i <= 10; i++) {
        Node *node = (Node *) malloc(sizeof(Node));
        node->value = i;
        node->next = NULL;
        p->next = node;
        p = node;
    }

    Node *list2_1 = (Node *) malloc(sizeof(Node));
    list2_1->next = NULL;
    p = list2_1;
    for (i = 11; i <= 20; i++) {
        Node *node = (Node *) malloc(sizeof(Node));
        node->value = i;
        node->next = NULL;
        p->next = node;
        p = node;
    }

    printf("原链表1：");
    printList(list1_1);
    printf("原链表2：");
    printList(list2_1);

    merge(list1_1, list2_1);

    printf("合并后的链表：");
    printList(list1_1);

    // 释放内存
    Node *curr = list1_1->next;
    while (curr != NULL) {
        Node *next = curr->next;
        free(curr);
        curr = next;
    }
    free(list1_1);

    curr = list2_1->next;
    while (curr != NULL) {
        Node *next = curr->next;
        free(curr);
        curr = next;
    }
    free(list2_1);

    // 第二组测试数据
    Node *list1_2 = (Node *) malloc(sizeof(Node));
    list1_2->next = NULL;
    p = list1_2;
    for (i = 1; i <= 10; i += 2) {
        Node *node = (Node *) malloc(sizeof(Node));
        node->value = i;
        node->next = NULL;
        p->next = node;
        p = node;
    }

    Node *list2_2 = (Node *) malloc(sizeof(Node));
    list2_2->next = NULL;
    p = list2_2;
    for (i = 2; i <= 10; i += 2) {
        Node *node = (Node *) malloc(sizeof(Node));
        node->value = i;
        node->next = NULL;
        p->next = node;
        p = node;
    }

    printf("原链表1：");
    printList(list1_2);
    printf("原链表2：");
    printList(list2_2);

    merge(list1_2, list2_2);

    printf("合并后的链表：");
    printList(list1_2);

    // 释放内存
    curr = list1_2->next;
    while (curr != NULL) {
        Node *next = curr->next;
        free(curr);
        curr = next;
    }
    free(list1_2);

    curr = list2_2->next;
    while (curr != NULL) {
        Node *next = curr->next;
        free(curr);
        curr = next;
    }
    free(list2_2);

    return 0;
}

输出结果：

原链表1：1 2 3 4 5 6 7 8 9 10 
原链表2：11 12 13 14 15 16 17 18 19 20 
合并后的链表：20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 
原链表1：1 3 5 7 9 
原链表2：2 4 6 8 10 
合并后的链表：10 9 8 7 6 5 4 3 2 1