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.

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Have you met this question in a real interview?

思路：这个利用2个指针p1,p2使他们之间的距离为n-1,一个指针p1提前走n-1，这样当右边的指针p1到最后一个元素的时候，左边的p2刚好到倒数第n个，所以最好还有一个辅助的指针pre指在p2的前一个元素，这样方便删除p1。注意当要删除第一个元素时，直接返回head.next即可。

代码如下：

/** * 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.next==null) return null;               ListNode pre = null;         ListNode p1 = head;        ListNode p2 = head;        while(n>1){            p2 = p2.next;            n--;        }                while(p2.next!=null){            pre = p1;            p1 = p1.next;            p2 = p2.next;                }                //要删除的是链表的第一个元素        if(pre == null){             return head.next;        }        pre.next = p1.next;                return head;            }}

画个图代码就比较容易得到了。