单向链表逆序

最新推荐文章于 2021-02-01 21:10:54 发布

原创最新推荐文章于 2021-02-01 21:10:54 发布 · 898 阅读

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#链表

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本文介绍了一种链表逆序的实现方法，并提供了一个简单的测试程序。通过三个关键步骤完成链表逆序：初始化指针、遍历并调整节点指向、返回新的头结点。

逆序函数：

NODE *reverse(NODE *head)
{
	NODE *p1, *p2, *temp;
	p1 = head;
	p2 = head->next;
	p1->next = NULL;
	while(p2 != NULL)
	{
		temp = p2->next;
		p2->next = p1;
		p1 = p2;
		p2 = temp;
	}
	return p1;
}

测试程序：

#include <stdio.h>
#include <stdlib.h>

typedef struct link
{
	int num;
	struct link *next;
}NODE;

NODE *create(int n)
{
	NODE *p, *head = NULL, *mov;
	int i = 0;
	for(i=0; i<n; i++)
	{
		p = (NODE *)malloc(sizeof(NODE));
		printf("please input num:\n");
		scanf("%d", &p->num);
		if(i == 0)
		{
			head = p;
			p->next = NULL;
		}
		else
		{
			mov = head;
			while(mov->next != NULL)
			{
				mov = mov->next;
			}
			mov->next = p;
			p->next = NULL;
		}
	}
	return head;
}

NODE *reverse(NODE *head)
{
	NODE *p1, *p2, *temp;
	p1 = head;
	p2 = head->next;
	p1->next = NULL;
	while(p2 != NULL)
	{
		temp = p2->next;
		p2->next = p1;
		p1 = p2;
		p2 = temp;
	}
	return p1;
}

void display(NODE *head)
{
	NODE *p = head;
	while(p->next != NULL)
	{
		printf("%d ", p->num);
		p = p->next;
	}
	printf("%d\n", p->num);
}

int main(int argc, char *argv[])
{
	NODE *head = NULL;

	head = create(5);
	display(head);

	head = reverse(head);
	display(head);

	return 0;
}