Sword finger offer merges two sorted linked lists

Title Description

Enter two monotonically increasing linked lists , Output the synthesized list of two linked lists , Of course, we need the synthesized list to satisfy the monotone rule .

 Solution1：

If one is empty , Then go back to the other . When neither is empty , Compare the size and choose the smaller one to add to the new linked list , Until one side is empty .

Finally, add the non empty linked list to the end of the new linked list .

Solution2:

The recursive method

Solution1:

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2){
		if (!pHead1)return pHead2;
		if (!pHead2)return pHead1;
		ListNode *p, *q, *head = nullptr;
		while (pHead1&&pHead2){
			if (pHead1->val < pHead2->val){
				p = pHead1;
				pHead1 = pHead1->next;
			}
			else{
				p = pHead2;
				pHead2 = pHead2->next;
			}
			if (head == nullptr){
				head = q = p;
			}
			else{
				q->next = p;
				q = p;
			}
		}
		if (pHead1)p->next = pHead1;
		if (pHead2)p->next = pHead2;
		return head;
	}
};

Solution2:

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        if(pHead1==null)return pHead2;
        if(pHead2==null)return pHead1;
        if(pHead1->val<pHead2->val){
            pHead1->next=Merge(pHead1->next,pHead2);
            return pHead1;
        }else{
            pHead2->next=Merge(pHead1,pHead2->next);
            return pHead2;
        }
    }
};