【链表&删除倒数第K个节点】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：删除链表的倒数第K个节点

解法：三个指针，p1指向尾节点，p2指向倒数第K个节点，pre指向要删除节点的前一个节点，注意特殊情况，刚好要删除的节点是头结点的情况

public class Solution {    public ListNode removeNthFromEnd(ListNode head, int n) {        ListNode p1 = head;        ListNode p2 = null, pre=null;        int k = 0;                while(p1 != null){            k++;            p1 = p1.next;                        if(k == n){                p2 = head;            }             else if(k > n){                pre = p2;                p2 = p2.next;            }        }                if(p2 != null){            if(pre == null) head = p2.next;            else pre.next = p2.next;        }                return head;    }}