[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.

思路：用三个指针，快指针、慢指针、指向慢指针前节点的指针，当快指针移动到链表最后一个节点时，慢指针指向倒数第k个节点，然后跳过第k个节点即可

代码：

/**
* 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) {
ListNode *preslow=NULL,*slow=head,*fast=head;
if(head==NULL) return head;
while((n-1)>0){
fast=fast->next;
n--;
}
if(fast->next==NULL) return head->next;
fast=fast->next->next;
slow=slow->next;
preslow=head;
while(fast!=NULL){
fast=fast->next;
slow=slow->next;
preslow=preslow->next;
}
if(slow==head){
return head->next;
}
preslow->next=slow->next;
return head;
}
};