链表删除节点

//倒序删除第n个结点
//链表的题目基本上都是优先考虑双指针。一个指针先走N步，
//然后两个指针同步移动到链表末尾，移除前一个指针所指着的节点即可 
#include<iostream>
using namespace std;
struct NODE
{
	int data;
	NODE *next;
};
NODE *Remove(NODE *L,int n) //*Remove的*不要忘记 
{
	NODE *head=L;
	NODE *right=head;
	NODE *left=head;
	int i;
	for(i=0;i<n;i++)
	{
		right=right->next;
	}
	if(right==NULL)
	{
		L=L->next;
		return head;
	}
	while(right->next!=NULL)
	{
		left=left->next;
		right=right->next;
	}
	left->next=left->next->next;
	return head; 
} 
int main()
{
	NODE s1,s2,s3,s4;
	s1.data=1;
	s1.next=&s2;
	s2.data=3;
	s2.next=&s3;
	s3.data=5;
	s3.next=&s4;
	s4.data=7;
	s4.next=NULL;
	int n=3;
	NODE *s=Remove(&s1,n);
	while(s!=NULL)
	{
		cout<<s->data<<endl;
		s=s->next;
	}
}