21. Merge Two Sorted Lists leetcode(lists)

Merge Two Sorted Lists
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
方法一、多开辟了一个链表

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {        if(NULL == l1 || NULL == l2)        {            return (NULL==l1)?l2:l1;        }        ListNode l3(0); //必须使用这种定义，如果写成ListNode* l3会超时，必须带头节点        ListNode* p1;        ListNode* p2;        ListNode* p3;        p1 = l1;        p2 = l2;        p3 = &l3;        while(p1 && p2)        {            ListNode* tmp = (ListNode*)malloc(sizeof(ListNode));            if(p1->val <= p2->val)            {                cout<<"1p1 "<<p1->val<<endl;                tmp->val = p1->val;                tmp->next = NULL;                p1 = p1->next;            }            else            {                cout<<"1p2 "<<p2->val<<endl;                tmp->val = p2->val;                tmp->next = NULL;                p2 = p2->next;            }            p3->next = tmp;            p3 = tmp;        }        while(p1)        {            cout<<"2p1"<<p1->val<<endl;            ListNode* tmp = (ListNode*)malloc(sizeof(ListNode));            tmp->val = p1->val;            tmp->next = NULL;            p3->next = tmp;            p3 = tmp;            p1 = p1->next;        }        while(p2)        {            cout<<"2p2"<<p2->val<<endl;            ListNode* tmp = (ListNode*)malloc(sizeof(ListNode));            tmp->val = p2->val;            tmp->next = NULL;            p3->next = tmp;            p3 = tmp;            p2 = p2->next;        }        return l3.next; //因为带头节点所以必须返回l3.next    }

方法二、直接在第一个链表上进行修改，最终返回第一个链表

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {        if(NULL == l1 || NULL == l2)        {            return (NULL==l1)?l2:l1;        }        ListNode* p1= l1;        ListNode* pre = p1;        ListNode* p2 = l2;        while(p1 && p2)        {            ListNode* tmp = (ListNode*)malloc(sizeof(ListNode));            if(p1->val <= p2->val)            {                pre = p1;                p1 = p1->next;            }            else            {                if(pre == p1) //注意当pre==p1时表示要将p2的节点插入p1的头节点                {                    tmp->val = p2->val;                    tmp->next = pre;                    pre = tmp;                    l1 = pre;                    p2 = p2->next;                }                else                {                    tmp->val = p2->val;                    pre->next = tmp;                    tmp->next = p1;                    p2 = p2->next;                    pre = tmp;                }            }        }        while(p2)        {            ListNode* tmp = (ListNode*)malloc(sizeof(ListNode));            tmp->val = p2->val;            tmp->next = NULL;            pre->next = tmp;            pre = tmp;            p2 = p2->next;        }        return l1;    }