合并两个排序链表

题目：将两个已经排序的单向链表合并为一个链表，要求空间复杂度尽可能的小。

思路：典型的归并排序思路，需要注意的两个地方时：1，怎样做到新链表不申请新的空间？2，代码尽量简洁？

下面是我的解答，希望跟各位网友一起讨论，算法本身是否有优化之处？如何能改进代码风格？

#include "stdafx.h"


class Node
{
public:
	int data;
	Node* next;
	Node() : data(-1), next(NULL){}
};


Node* Merge (Node* head1, Node* head2)
{
	Node* head = NULL;
	Node* ret = NULL;
	while (head1 || head2)
	{
		Node* temp = NULL;


		if (head1 && head2)
		{
			temp = head1->data < head2->data ? head1 : head2;
		}
		else 
		{
			temp = head1 !=  NULL ? head1 : head2;
		}




		if (!head)
		{
			head = temp;
			ret = head;
		}
		else
		{
			head->next = temp;
			head = head->next;
		}


		if (temp == head1)
		{
			head1 = head1->next;
		}
		else
		{
			head2 = head2->next;
		}
		
	}
	head->next = NULL;
	return ret;
}


void Print(Node* head)
{
	while(head)
	{
		printf("%d ", head->data);
		head = head->next;
	}
	printf("\n");
}


void Test0()
{
	Node* node1 = new Node();
	Node* node2 = new Node();
	printf("before merge\n");
	Print(node1);
	Print(node2);
	Node* head = Merge(node1, node2);
	printf("after merge\n");
	Print(head);
}


void Test1()
{
	Node* node1 = new Node();
	Node* head1 = node1;
	node1->data = 0;
	Node* node2 = new Node();
	Node* head2 = node2;
	node2->data = 1;


	for (int i = 2; i < 6; i+=2)
	{
		node1->next = new Node();
		node1 = node1->next;
		node1->data = i;
		node1->next = NULL;


		node2->next = new Node();
		node2 = node2->next;
		node2->data = i + 1;
		node2->next = NULL;
	}
	printf("before merge\n");
	Print(head1);
	Print(head2);
	Node* head = Merge(head1, head2);
	printf("after merge\n");
	Print(head);


}


void Test2()
{
	Node* node1 = new Node();
	Node* head1 = node1;
	node1->data = 0;
	Node* node2 = new Node();
	Node* head2 = node2;
	node2->data = 1;


	for (int i = 2; i <= 6; i+=2)
	{
		node1->next = new Node();
		node1 = node1->next;
		node1->data = i;
		node1->next = NULL;


		node2->next = new Node();
		node2 = node2->next;
		node2->data = i + 1;
		node2->next = NULL;
	}


	for (int j = 8; j <= 10; j++)
	{
		node1->next = new Node();
		node1 = node1->next;
		node1->data = j;
		node1->next = NULL;
	}
	printf("before merge\n");
	Print(head1);
	Print(head2);
	printf("after merge\n");
	Node* head = Merge(head1, head2);
	Print(head);


}


void Test3()
{
	Node* node1 = new Node();
	Node* head1 = node1;
	node1->data = 0;
	Node* node2 = new Node();
	Node* head2 = node2;
	node2->data = 1;


	for (int i = 2; i <= 6; i+=2)
	{
		node1->next = new Node();
		node1 = node1->next;
		node1->data = i;
		node1->next = NULL;


		node2->next = new Node();
		node2 = node2->next;
		node2->data = i + 1;
		node2->next = NULL;
	}


	for (int j = 7; j <= 10; j++)
	{
		node2->next = new Node();
		node2 = node2->next;
		node2->data = j;
		node2->next = NULL;
	}
	printf("before merge\n");
	Print(head1);
	Print(head2);
	Node* head = Merge(head1, head2);
	printf("after merge\n");
	Print(head);


	//head1, head2 should do the memory collection job. skip for testing reason.
}






int _tmain(int argc, _TCHAR* argv[])
{
	Test0();
	Test1();
	Test2();
	Test3();
	//should be more test like negative value, integer overflow...skip 
	return 0;
}