练手---＞两个有序单向链表合并C/C++

1. 创建一个新的链表
2. 进行循环比较，将较小值赋值给新链表的新节点，注意：每次都要创建新的节点，开辟空间，将较小值赋值给该节点，再将该节点尾插在新建链表中

//2020_10_24
//两个有序单向链表合并
#include<isotream>

using namespace std;
struct Node_d {
	int data;
	Node_d* next;
};
 Node_d* mixList( Node_d* head1, Node_d* head2) 
{
	if (head1 == nullptr)
		return head2;
	else if(head2 == nullptr)
		return head1;
	Node_d* newhead;
	if (head1->data <= head2->data)
	{
		newhead = head1;
		head1 = head1->next;
	}
	else
	{
		newhead = head2;
		head2 = head2->next;
	}
	Node_d* temp = newhead;
	while (head1 != NULL && head2 != NULL)
	{
		Node_d* cur = new Node_d;
		if (head1->data <= head2->data)
		{
			cur->data = head1->data;
			temp->next = cur;
			head1 = head1->next;
		}
		else
		{
			cur->data = head2->data;
			temp->next = cur;
			head2 = head2->next;
		}
		temp = temp->next;
	}
	if (head1 == nullptr)
		temp->next = head2;
	else
		temp->next = head1;
	return newhead;
}

void testmixList()
{
	Node_d* head1=new Node_d;
	Node_d* head2=new Node_d;
	head1->data = 0;
	head2->data = 0;
	Node_d* temp1=head1;
	Node_d* temp2 = head2;

	for (int i = 1; i < 5; ++i)
	{
		Node_d* cur=new Node_d;
		cur->data = i;
		temp1->next = cur;
		temp2->next = cur;
		temp1 = temp1->next;
		temp2 = temp2->next;
		temp1->next = nullptr;
		temp2->next = nullptr;
	}
	Node_d* newhead = mixList(head1, head2);

	Node_d* temp = newhead;
	while (temp != nullptr)
	{
		if (temp != nullptr)
		{
			cout << temp->data << " ";
			temp = temp->next;
		}
	}
	cout << endl;
}
}

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