将两个非递减单链表合并成一个有序的非递减单链表的c语言代码，利用头插法

以下是将两个非递减单链表合并成一个有序的非递减单链表的c语言代码，利用头插法：

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

// 定义链表节点结构体
typedef struct ListNode {
    int val;
    struct ListNode* next;
} ListNode;

// 头插法合并两个非递减单链表
ListNode* mergeLists(ListNode* l1, ListNode* l2) {
    ListNode* dummy = (ListNode*)malloc(sizeof(ListNode)); // 虚拟头节点
    dummy->next = NULL;
    ListNode* cur = dummy;
    while (l1 && l2) {
        if (l1->val <= l2->val) {
            cur->next = l1;
            l1 = l1->next;
        } else {
            cur->next = l2;
            l2 = l2->next;
        }
        cur = cur->next;
    }
    cur->next = l1 ? l1 : l2; // 将剩余链表连接到已排序的链表
    return dummy->next;
}

int main(void) {
    // 创建链表1：1 -> 3 -> 5 -> 7 -> 9
    ListNode* l1 = (ListNode*)malloc(sizeof(ListNode));
    l1->val = 1;
    ListNode* l1_2 = (ListNode*)malloc(sizeof(ListNode));
    l1_2->val = 3;
    l1->next = l1_2;
    ListNode* l1_3 = (ListNode*)malloc(sizeof(ListNode));
    l1_3->val = 5;
    l1_2->next = l1_3;
    ListNode* l1_4 = (ListNode*)malloc(sizeof(ListNode));
    l1_4->val = 7;
    l1_3->next = l1_4;
    ListNode* l1_5 = (ListNode*)malloc(sizeof(ListNode));
    l1_5->val = 9;
    l1_4->next = l1_5;
    l1_5->next = NULL;

    // 创建链表2：2 -> 4 -> 6 -> 8 -> 10
    ListNode* l2 = (ListNode*)malloc(sizeof(ListNode));
    l2->val = 2;
    ListNode* l2_2 = (ListNode*)malloc(sizeof(ListNode));
    l2_2->val = 4;
    l2->next = l2_2;
    ListNode* l2_3 = (ListNode*)malloc(sizeof(ListNode));
    l2_3->val = 6;
    l2_2->next = l2_3;
    ListNode* l2_4 = (ListNode*)malloc(sizeof(ListNode));
    l2_4->val = 8;
    l2_3->next = l2_4;
    ListNode* l2_5 = (ListNode*)malloc(sizeof(ListNode));
    l2_5->val = 10;
    l2_4->next = l2_5;
    l2_5->next = NULL;

    // 合并两个链表
    ListNode* merged = mergeLists(l1, l2);

    // 遍历输出合并后的链表
    printf("合并后的链表：");
    ListNode* node = merged;
    while (node) {
        printf("%d ", node->val);
        node = node->next;
    }
    printf("\n");

    // 释放链表节点的内存
    node = merged;
    while (node) {
        ListNode* tmp = node;
        node = node->next;
        free(tmp);
    }
    node = l1;
    while (node) {
        ListNode* tmp = node;
        node = node->next;
        free(tmp);
    }
    node = l2;
    while (node) {
        ListNode* tmp = node;
        node = node->next;
        free(tmp);
    }

    return 0;
}