Leetcode 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.

题意转换一下就是求解单链表的倒数第k个节点，解法是设置两个指针，一个在前一个在后，之间的距离为k，同时向前移动，有点类似滑动窗口，当第一个指针到达链表尾的时候，第二个指针即为倒数第k个节点，具体的见编程之美

ListNode *removeNthFromEnd(ListNode *head, int n){
if(head == NULL) return NULL;
ListNode *q = head,*p = head;
for(int i = 0 ; i < n - 1; ++ i)
q = q -> next;
ListNode *pPre = NULL;
while(q->next){
pPre = p;
p = p->next;
q = q->next;
}
if(pPre == NULL){
pPre = p->next;
delete p;
}else{
pPre ->next = p ->next;
delete p;
}
return head;
}