删除链表的倒数第n个结点 Remove Nth Node From End of List

题目源自于Leetcode。

题目：Given a linked list, remove the nth node
from the end of list and return its head.只允许过一遍。

思路：注意当原链表节点个数小于n和等于n的时候这两种特殊情况。

/**
* 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) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if(n<=0)
return head;
ListNode *pre, *p;
pre = p = head;
int i;
for(i=0;i<n;i++)
if(p->next!=NULL)
p = p->next;
else
break;
if(i<n-1)//原链表总个数小于n个
return head;
if(i == n-1)//原链表总个数等于n，即要删除链首结点
{
ListNode *tmp1 = head;
ListNode *tmp2 = head->next;
delete tmp1;
return tmp2;
}
while(p->next!=NULL)
{
p = p->next;
pre = pre->next;
}
ListNode *tmp = pre->next;
pre->next = tmp->next;
delete tmp;
return head;
}
};