用C语言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 ListNode {
    int val;
    struct ListNode* next;
};

struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    if (l1 == NULL) {
        return l2;
    }
    if (l2 == NULL) {
        return l1;
    }
    struct ListNode* head = NULL;
    struct ListNode* tail = NULL;
    while (l1 != NULL && l2 != NULL) {
        if (l1->val < l2->val) {
            if (head == NULL) {
                head = tail = l1;
            }
            else {
                tail->next = l1;
                tail = l1;
            }
            l1 = l1->next;
        }
        else {
            if (head == NULL) {
                head = tail = l2;
            }
            else {
                tail->next = l2;
                tail = l2;
            }
            l2 = l2->next;
        }
    }
    if (l1 != NULL) {
        tail->next = l1;
    }
    if (l2 != NULL) {
        tail->next = l2;
    }
    return head;
}

int main() {
    struct ListNode* l1 = (struct ListNode*)malloc(sizeof(struct ListNode));
    l1->val = 1;
    l1->next = (struct ListNode*)malloc(sizeof(struct ListNode));
    l1->next->val = 3;
    l1->next->next = (struct ListNode*)malloc(sizeof(struct ListNode));
    l1->next->next->val = 5;
    l1->next->next->next = NULL;

    struct ListNode* l2 = (struct ListNode*)malloc(sizeof(struct ListNode));
    l2->val = 2;
    l2->next = (struct ListNode*)malloc(sizeof(struct ListNode));
    l2->next->val = 4;
    l2->next->next = (struct ListNode*)malloc(sizeof(struct ListNode));
    l2->next->next->val = 6;
    l2->next->next->next = NULL;

    struct ListNode* merged = mergeTwoLists(l1, l2);
    while (merged != NULL) {
        printf("%d ", merged->val);
        merged = merged->next;
    }
    printf("\n");

    return 0;
}