Remove Nth Node From End of List --移除链表中的倒数第k个元素

问题：链接

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.

解答：
注意如果移除的元素是头元素的话，需要特别的处理，否则的话直接找到倒数第n+1个元素，移除它后面的元素即可。

代码：

class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {
int len=0;
ListNode *pf = head;
while(pf != NULL)
{
++len;
pf = pf->next;
}
if(len == n)
return head->next;
pf = head;
ListNode *pl = head;
for(int i = 1; i <= n && pf->next != NULL; i++)
pf = pf->next;
while(pf->next != NULL)
{
pf = pf->next;
pl = pl->next;
}
pl->next = pl->next->next;
return head;
}
};