用C代码实现将两个递增的有序链表合并为一个递增的有序链表

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

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

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

struct Node* mergeLists(struct Node* list1, struct Node* list2) {
    struct Node* mergedList = createNode(0);
    struct Node* tail = mergedList;

    while (1) {
        if (list1 == NULL) {
            tail->next = list2;
            break;
        }
        else if (list2 == NULL) {
            tail->next = list1;
            break;
        }

        if (list1->data <= list2->data) {
            tail->next = list1;
            list1 = list1->next;
        }
        else {
            tail->next = list2;
            list2 = list2->next;
        }

        tail = tail->next;
    }

    return mergedList->next;
}

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

int main() {
    struct Node* list1 = createNode(1);
    list1->next = createNode(3);
    list1->next->next = createNode(5);
    list1->next->next->next = createNode(7);

    struct Node* list2 = createNode(2);
    list2->next = createNode(4);
    list2->next->next = createNode(6);
    list2->next->next->next = createNode(8);

    printf("List 1: ");
    printList(list1);

    printf("List 2: ");
    printList(list2);

    struct Node* mergedList = mergeLists(list1, list2);

    printf("Merged List: ");
    printList(mergedList);

    return 0;
}

输出:

List 1: 1 3 5 7
List 2: 2 4 6 8
Merged List: 1 2 3 4 5 6 7 8