Leetcode Remove Nth Node From End of List 删除链表倒数第n个元素

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.

分析：

1. 删除的可能是正数第一个元素，因此需要设置dummy node。

2. 设置两个指针，快指针先走n-1步，然后快慢指针一起向前移动，一次动一步。

3. 由于找到需要删除的点后需要把它前面节点的next设为要删除节点的next，所以需要记录慢指针前面的节点。

Java代码实现：

/** * 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)            return head;                    ListNode dummy = new ListNode(0);        dummy.next = head;        ListNode fast = head;        ListNode slow = head;        ListNode pre = dummy;                for(int i=0;i<n-1;i++)        {            fast = fast.next;        }        while(fast.next!=null)        {            fast = fast.next;            slow = slow.next;              pre = pre.next;        }        pre.next = slow.next;                return dummy.next;    }}