用c语言链表实现A={7,5,3,11}何B=｛2,6,3｝和并｝

以下是用C语言链表实现A={7,5,3,11}和B={2,6,3}合并的代码：

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

typedef struct Node {
    int data;
    struct Node* next;
} Node;

Node* createNode(int data) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}

void printList(Node* head) {
    Node* current = head;
    while (current != NULL) {
        printf("%d ", current->data);
        current = current->next;
    }
    printf("\n");
}

Node* mergeLists(Node* list1, Node* list2) {
    Node* head = NULL;
    Node* tail = NULL;
    while (list1 != NULL && list2 != NULL) {
        if (list1->data < list2->data) {
            if (head == NULL) {
                head = list1;
                tail = head;
            } else {
                tail->next = list1;
                tail = tail->next;
            }
            list1 = list1->next;
        } else {
            if (head == NULL) {
                head = list2;
                tail = head;
            } else {
                tail->next = list2;
                tail = tail->next;
            }
            list2 = list2->next;
        }
    }
    if (list1 != NULL) {
        if (head == NULL) {
            head = list1;
        } else {
            tail->next = list1;
        }
    }
    if (list2 != NULL) {
        if (head == NULL) {
            head = list2;
        } else {
            tail->next = list2;
        }
    }
    return head;
}

int main() {
    Node* A = createNode(7);
    A->next = createNode(5);
    A->next->next = createNode(3);
    A->next->next->next = createNode(11);

    Node* B = createNode(2);
    B->next = createNode(6);
    B->next->next = createNode(3);

    printf("A: ");
    printList(A);
    printf("B: ");
    printList(B);

    Node* mergedList = mergeLists(A, B);
    printf("Merged list: ");
    printList(mergedList);

    return 0;
}

输出：

A: 7 5 3 11 
B: 2 6 3 
Merged list: 2 6 3 7 5 3 11