Remove Nth Node From End of List

题目：

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:

Given n will always be valid.

Try to do this in one pass.

思路1：

先将链表逆转，然后删除相应位的元素，再次逆转，输出链表

思路2：

设定两个指针，二者相差n-1，当后面的一个指针为空时，前一个指针即为所要删除的节点

代码：

#include <iostream>

using namespace std;

/**
* Definition for singly-linked list. */
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) { //思路1
/*将链表逆转*/
if(head==NULL) return head;//head为空直接返回
if(head->next==NULL && n==1) //链表只包含一个元素，直接删除
{
head=NULL;
return head;
}
ListNode *q; //用于记录头结点
ListNode *nex;//下一个节点
ListNode *pre=NULL;//前一个节点
ListNode *cur=head;//当前节点
while(cur!=NULL)
{
nex=cur->next;
cur->next=pre;
pre=cur;
cur=nex;
}
head=pre;

/*删除特定位置上的值*/
q=head;
if(n==1)
{
head=head->next;
}
else
{
while(--n)
{
pre=q;
q=q->next;
}
pre->next=q->next;
}

/*再次将链表逆转*/
pre=NULL;
cur=head;
while(cur!=NULL)
{
nex=cur->next;
cur->next=pre;
pre=cur;
cur=nex;
}
head=pre;

return head;
}

ListNode* removeNthFromEnd2(ListNode* head, int n) //思路2
{
if(head==NULL) return head;

ListNode *first,*last,*pre=NULL;
first=last=head;
n--;
while(n--&&last!=NULL)
{
last=last->next;
}

while(last->next!=NULL)
{
pre=first;
first=first->next;
last=last->next;
}
if(pre ==NULL)
{
head=first->next;
delete first;
}
else
{
pre->next=first->next;
delete first;
}

return head;
}
};

int main()
{
ListNode *head=new ListNode(1);
ListNode *p;
p=head;
/*
p->next=new ListNode(2);
p=p->next;

p->next=new ListNode(3);
p=p->next;

p->next=new ListNode(4);
p=p->next;

p->next=new ListNode(5);
p=p->next;
*/
Solution s;
head=NULL;
head=s.removeNthFromEnd2(head,1);

while(head!=NULL)
{
cout<<head->val<<"->";
head=head->next;
}

system("pause");
return 0;
}