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.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

// Merge_Two_Sorted_Lists.cpp: 定义控制台应用程序的入口点。
#include "stdafx.h"
#include<stdio.h>
#include<malloc.h>
struct ListNode {
	int val;
	struct ListNode *next;
};

struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
	struct ListNode* p, *L;
	p = (struct ListNode *)malloc(sizeof(struct ListNode));
	p->next = NULL;
	L = p;
	while (l1 != NULL || l2 != NULL) {
		if (l1 == NULL) {
			p->next = l2;
			break;
		}
		else if (l2 == NULL) {
			p->next = l1;
			break;
		}
		if (l1->val >= l2->val) {
			p->next = l2;
			l2 = l2->next;
			p = p->next;
			p->next = NULL;
		}
		else {
			p->next = l1;
			l1 = l1->next;
			p = p->next;
			p->next = NULL;
		}
	}
	return L->next;
}

int main()
{
	struct ListNode *l1,*l2,*l3;
	l1= (struct ListNode *)malloc(sizeof(struct ListNode));   //申请头结点空间
	l1->next = NULL;                  //初始化一个空链表
	l1->val = 1;
	l1->next = NULL;
	l2 = (struct ListNode *)malloc(sizeof(struct ListNode));   //申请头结点空间
	l2->val = 2;
	l2->next = l1->next;
	l1->next = l2;
	l3 = (struct ListNode *)malloc(sizeof(struct ListNode));
	l3->val = 4;
	l3->next = l1->next->next;
	l1->next->next = l3;
	

	struct ListNode *p1, *p2, *p3;
	p1 = (struct ListNode *)malloc(sizeof(struct ListNode));   //申请头结点空间
	p1->next = NULL;                  //初始化一个空链表
	p1->val = 1;
	p1->next = NULL;
	p2 = (struct ListNode *)malloc(sizeof(struct ListNode));   //申请头结点空间
	p2->val = 3;
	p2->next =p1->next;
	p1->next = p2;
	p3 = (struct ListNode *)malloc(sizeof(struct ListNode));
	p3->val = 4;
	p3->next =p1->next->next;
	p1->next->next = p3;
	
	//struct ListNode l2;
	struct ListNode *L = mergeTwoLists(l1, p1);
	while (L != NULL) {
		printf("%d\n", L->val);
		L = L->next;
	}
    return 0;
}