合并两个已排序的链表

上述uml图下载地址：http://download.youkuaiyun.com/detail/bladeandmaster88/9631377

#include <iostream>
using namespace std;

struct Node
{
	Node(int x)
	{
		value = x;
		next = 0;
	}
	int value;
	Node *next;
};


Node *Merge(Node *head1, Node *head2)
{
	if (head1 == NULL)
		return head2;
	if (head2 == NULL)
		return head1;

	Node *head = NULL;//新链表的头节点

	//确定新链表的头节点
	if (head1->value <= head2->value)
	{
		head = head1;
		head1 = head1->next;
	}
	else
	{
		head = head2;
		head2 = head2->next;
	}

	Node *cur = head;//新链表的当前节点

	//在两个链表中遍历比较，将值较小的结点链接到pcur结点后
	while ( head1 && head2)
	{
		if (head1->value <= head2->value)
		{
			cur->next = head1;
			cur=head1;
			head1 = head1->next;
		}
		else
		{
			cur->next = head2;
			cur = head2;
			head2 = head2->next;
		}
	}

	//将head1或head2剩余的结点链到cur之后，完成整个合并的过程
	if (head1 )//对于这个例子，head1为空
		cur->next = head1;

	if (head2)
		cur->next = head2;

	return head;
}

int main()
{
	Node *p1 = new Node(1);
	Node *p2 = new Node(3);
	Node *p3 = new Node(5);
	Node *p4 = new Node(7);
	Node *p5 = new Node(9);
	Node *p6 = new Node(11);

	p1->next = p2;
	p2->next = p3;
	p3->next = p4;
	p4->next = p5;
	p5->next = p6;

	Node *q1 = new Node(2);
	Node *q2 = new Node(4);
	Node *q3 = new Node(6);
	Node *q4 = new Node(8);
	Node *q5 = new Node(10);
	Node *q6 = new Node(12);

	q1->next = q2;
	q2->next = q3;
	q3->next = q4;
	q4->next = q5;
	q5->next = q6;

	Node *ptr =0;
	ptr = Merge(p1, q1);

	while (ptr)
	{
		cout<<ptr->value<<" ";
		ptr = ptr->next;
	}
	cout << endl;

	return 0;
}