代码实现：

方法1：递归  ---->难点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* mergeTwoLists(struct ListNode *list1, struct ListNode *list2) {
    /*
        1.如果l1为空，返回l2
        2.如果l2为空，返回l1
        3.如果l1的值小于l2，比较l1的next值和l2，并把值赋给l1的下一个；返回l1
        4.反之，比较l1和l2的next值，并把值赋给l2的下一个；返回l2
    */
    if (list1 == NULL) {
        return list2;
    } else if (list2 == NULL) { 
        return list1;
    }
    if (list1->val < list2->val) { 
        list1->next = mergeTwoLists(list1->next, list2);
        return list1;
    } else {
        list2->next = mergeTwoLists(list1, list2->next);
        return list2;
    }
}

方法2：常规解法+设置虚拟头结点                                                                           

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* mergeTwoLists(struct ListNode *list1, struct ListNode *list2) {
    if (list1 == NULL) {
        return list2;
    }
    if (list2 == NULL) {
        return list1;
    }
    struct ListNode *head = malloc(sizeof(*head)); // 设置虚拟头结点
    struct ListNode *h = head;
    while (list1 && list2) {
        if (list1->val < list2->val) {
            h->next = list1;
            list1 = list1->next;
        } else {
            h->next = list2;
            list2 = list2->next;
        }
        h = h->next;
        h->next = NULL;
    }
    if (list1) {
        h->next = list1;
    }
    if (list2) {
        h->next = list2;
    }
    struct ListNode *result = head->next;
    head->next = NULL;
    free(head);
    return result;    
}