用C语言写一个算法，已知两个链表A和B分别表示两个集合，其元素递增排列。请设计一个算法，用户求出A和B的差集（即仅由在A中出现而在B中不出现的元素构成的集合），并用递增的方式保存在表A中，同时返回该集合的元素数目

以下是算法的C语言实现：

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode* createList(int* nums, int size) {
    struct ListNode* head = NULL;
    struct ListNode* tail = NULL;
    for (int i = 0; i < size; i++) {
        struct ListNode* node = (struct ListNode*)malloc(sizeof(struct ListNode));
        node->val = nums[i];
        node->next = NULL;
        if (head == NULL) {
            head = node;
            tail = node;
        } else {
            tail->next = node;
            tail = node;
        }
    }
    return head;
}

void printList(struct ListNode* head) {
    while (head != NULL) {
        printf("%d ", head->val);
        head = head->next;
    }
    printf("\n");
}

struct ListNode* subtractLists(struct ListNode* A, struct ListNode* B, int* size) {
    struct ListNode* head = NULL;
    struct ListNode* tail = NULL;
    int count = 0;
    while (A != NULL && B != NULL) {
        if (A->val < B->val) {
            struct ListNode* node = (struct ListNode*)malloc(sizeof(struct ListNode));
            node->val = A->val;
            node->next = NULL;
            if (head == NULL) {
                head = node;
                tail = node;
            } else {
                tail->next = node;
                tail = node;
            }
            A = A->next;
            count++;
        } else if (A->val == B->val) {
            A = A->next;
            B = B->next;
        } else {
            B = B->next;
        }
    }
    while (A != NULL) {
        struct ListNode* node = (struct ListNode*)malloc(sizeof(struct ListNode));
        node->val = A->val;
        node->next = NULL;
        if (head == NULL) {
            head = node;
            tail = node;
        } else {
            tail->next = node;
            tail = node;
        }
        A = A->next;
        count++;
    }
    *size = count;
    return head;
}

int main() {
    int nums1[] = {1, 2, 3, 4, 5};
    int nums2[] = {3, 4, 5, 6, 7};
    struct ListNode* A = createList(nums1, 5);
    struct ListNode* B = createList(nums2, 5);
    int size;
    struct ListNode* C = subtractLists(A, B, &size);
    printf("A - B = ");
    printList(C);
    printf("size = %d\n", size);
    return 0;
}

输出结果为：

A - B = 1 2 
size = 2

其中，函数`createList`用于创建链表，函数`printList`用于打印链表，函数`subtractLists`用于求两个链表的差集，函数`main`用于测试。