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l22.如果l2为空，返回l13.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l14.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;    
}