LeetCode(46)-Remove Nth Node From End of List

题目：

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：

题意：这道题是给定一个链表，给定一个数字n，然后倒数去掉第n个数字，返回head

利用双指针，一个first先走n个，然后next在和first同步走，等到first.next ！= null；这时next正好指向要去掉元素的前一个，next.next = next.next.next;

注意去掉head的情况和链表为0和1

代码：

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
if(head == null||head.next == null){
return null;
}
ListNode prev = head;
ListNode next = head;
for(int i = 0;i < n;i++){
prev = prev.next;
}
if(prev == null){
head = head.next;
return head;
}
while(prev.next != null){
prev = prev.next;
next = next.next;
}
next.next = next.next.next;
return head;
}
}