Leetcode-19: 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.

试着遍历一遍链表，把倒数第N个节点删除掉。

思路：保证列表的长度不超过N；使用两个指针fast和slow遍历列表，fast先走N步，然后slow和fast一起往前走，这样slow落后fast N步，当fast走到最后一个节点时，slow在倒数第N个节点之前。为了方便，我们创建一个哨兵节点作为辅助。

/** * 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 || n <= 0) return head;        ListNode sential = new ListNode(0);        sential.next = head;        ListNode fast = sential, slow = sential;        for (int i = 0; i < n; ++i) {            if (fast == null)                return null;            fast = fast.next;        }        while (fast != null && fast.next != null) {            fast = fast.next;            slow = slow.next;        }        slow.next = slow.next.next;        return sential.next;    }}