一个简单的链表实现

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

typedef struct student
{
	int data;
	struct student* next;	
}node;


node* create()
{
	node*head,*p,*s;
	int x,cycle = 1;
	
	head = (node*)malloc(sizeof(node));
	p = head;
	while(cycle)
	{
		cout << "输入数据: ";
		cin >> x;
		if(x!=0)
		{
			s = (node*)malloc(sizeof(node));
			s->data = x;
			s->next = NULL;
			p->next = s;
			p = s;	
		}
		else
		{
			cycle = 0;
		}
	}
	p = head;	 
	head = head->next;
	p->next = NULL;
	delete p;
	s =  NULL;
	delete s;
	return head;
}

//求链表长度 
int len(node* head)
{
	int count = 0;
	node* p = head;
	while(p!=NULL)
	{
		++count;
		p=p->next;
	}
	return count;
}

//打印链表元素 
void print(node* head) 
{
	node* p = head;
	while(p!=NULL)
	{
		cout << p->data << " "; 
		p=p->next;
	}	
	cout << endl;
}
//链表排序 
node* sort(node* head) 
{
	//链表为空或者只有一个元素 
	if(head==NULL || head->next==NULL)	
	{
		return head;
	}
	
	int n = len(head); 
	node* p =NULL;
	for(int i=1; i<n; ++i)
	{
		p = head;
		for(int j=0; j<n-i;++j)
		{
			if(p->data > p->next->data)
			{
				int temp = p->data;
				p->data = p->next->data;
				p->next->data = temp;
			}
			p = p->next;
		}		
	}
	return head;
}

//链表逆置 
node* reverse(node* head)
{
	if(head==NULL||head->next==NULL)
	{
		return head;
	}

	node* p1;
	node* p2;
	node* p3;
	p1 = head;
	p2 = p1->next;
	while(p2!=NULL)
	{
		p3 = p2->next;
		p2->next = p1;
		p1 = p2;
		p2 = p3;	
	}
	head->next = NULL;
	head = p1;
	return head;
} 


int main()
{
	node* head = create();
	cout << endl;
	cout << "元素长度为: " << len(head) << endl; 
	cout << "排序前各个元素为: "; 
	print(head);
	
	cout << "排序后各个元素为: "; 
	head = sort(head); 
	print(head);
	
	cout << "逆置后各个元素为: "; 
	head = reverse(head); 
	print(head);
	
	return 0;
}