编程实现合并两个有序单链表

最新推荐文章于 2021-06-25 21:46:17 发布

tianmo2010

最新推荐文章于 2021-06-25 21:46:17 发布

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分类专栏：链表文章标签：编程 null merge struct list p2p

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本文介绍了一种在不使用额外内存的情况下合并两个降序链表的方法，并提供了两种实现方案：迭代和递归。通过示例代码展示了如何创建链表、合并链表以及显示合并后的链表。

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编程实现合并两个有序(假定为降序)单链表的函数，输入为两个有序链表的头结点，函数返回合并后新的链表的头节点，

要求：不能另外开辟新的内存存放合并的链表。

合并函数声明如下：node *merge_sorted_list( node * head1, node *head2);

//重要方法：

node merge_sorted_list(const node head1,const node head2)
{
	if((NULL == head1) && (NULL == head1))
	{
		return NULL;
	}
	else if(NULL == head1)
	{
		return head2;
	}
	else if(NULL == head2)
	{
		return head1;
	}
	else
	{	
		node head = NULL,p1 = NULL,p2 = NULL;


		if(head1->value >= head2->value)
		{
			head = head1;
			p1 = head1->next;
			
			p2 = head2;
		}
		else
		{		
			head = head2;
			p2 = head2->next;
			
			p1 = head1;
		}

		node p = head;

		while((NULL != p1) && (NULL != p2))
		{
			if(p1->value >= p2->value)
			{
			
				p->next = p1;


				p = p1;


				p1 = p1->next;
			}
			else
			{
				
				p->next = p2;


				p = p2;


				p2 = p2->next;
			}
		}
		
		p->next = p1 ? p1 : p2;
		//if(NULL != p1->next)
		//	p->next = p1;
		//else
		//	p->next = p2;
		return head;		
	}
}

采用递归的方法实现：

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

	if ( head2 == NULL)
		return head1 ;

	Node *head = NULL ;

	if ( head1->value > head2->value )
	{
		head = head1 ;
		head->next = MergeRecursive(head1->next,head2);
	}
	else
	{
		head = head2 ;
		head->next = MergeRecursive(head1,head2->next);
	}

	return head ;
}

//掌握好了吗？

//不带头节点的链表
#include <stdio.h>


#include <malloc.h>


struct Node
{
	int value;
	struct Node *next;
};


typedef struct Node * node;


node CreateList(int iStart,int num)
{
	node head = NULL,p = NULL,q = NULL;
	
	for(int i = num;i > 0 ;i --)
	{
		p = (node)malloc(sizeof(struct Node));
		p->value = iStart + i;


		if(head == NULL)
		{
			head = p;
		}
		else
		{
			q->next = p;
		}
		q = p;		
	}


	p->next = NULL;


	return head;
}


node merge_sorted_list(node head1,node head2)
{
	if((NULL == head1) && (NULL == head2))
	{
		return NULL;
	}
	else if(NULL == head1)
	{
		return head2;
	}
	else if(NULL == head2)
	{
		return head1;
	}
	else
	{
		node p = NULL,p1 = NULL,p2 = NULL;


		if(head1->value >= head2->value)
		{
			p = head1;
			p1 = head1->next;
			
			p2 = head2;
		}
		else
		{		
			p = head2;
			p2 = head2->next;
			
			p1 = head1;
		}


		while((NULL != p1) && (NULL != p2))
		{
			if(p1->value >= p2->value)
			{
			
				p->next = p1;


				p = p1;


				p1 = p1->next;
			}
			else
			{
				
				p->next = p2;


				p = p2;


				p2 = p2->next;
			}
		}	
		p->next = p1 ? p1 : p2;


		return head1->value >= head2->value ? head1 : head2;
	}
}


void display(node head)
{
	node p = head;


	while(NULL != p)
	{
		printf("%3d",p->value);


		p = p->next;
	}
	printf("\n");
}


int main()
{
	node head1 = NULL ,head2 = NULL,head = NULL;


	head1 = CreateList(5,6);
	head2 = CreateList(5,6);


	display(head1);
	display(head2);


	head = merge_sorted_list(head1,head2);
	display(head);


	return 0;


}