Remove Nth Node From End of List

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.

由于删除头结点时要改变头结点的指针，所以传入的是引用或者使用二级指针。提示中给的只是传入一级指针，要是删除头结点，始终不能删掉。（注意要改变一个数据结构时，要传二级指针或者引用，树中的很多操作都是这样的）

C++实现代码如下：

#include<iostream>
#include<new>
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)
{
if(head==NULL)
return NULL;
ListNode *p=head;
ListNode *q=head;
ListNode *pre=head;
int i=0;
while(q&&i<n)
{
q=q->next;
i++;
}
//说明没有n个元素
if(i<n)
return head;
while(q)
{
pre=p;
p=p->next;
q=q->next;
}
if(p==head)
{
head=head->next;
delete p;
}
else
{
pre->next=p->next;
delete p;
}
return head;
}
void createList(ListNode *&head)
{
ListNode *p=NULL;
int i=0;
int arr[10]= {10,9,8,7,6,5,4,3,2,1};
for(i=0; i<5; i++)
{
if(head==NULL)
{
head=new ListNode(arr[i]);
if(head==NULL)
return;
}
else
{
p=new ListNode(arr[i]);
p->next=head;
head=p;
}
}
}
};

int main()
{
Solution s;
ListNode *L=NULL;
s.createList(L);
ListNode *head=L;
while(head)
{
cout<<head->val<<" ";
head=head->next;
}
cout<<endl;
s.removeNthFromEnd(L,5);
while(L)
{
cout<<L->val<<" ";
L=L->next;
}
}

运行结果：


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