leetcode019 Remove Nth Node From End of List

题目

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.

思路：

利用很简单的数据结构知识就可以做出来 。先遍历一遍，找出链表的长度，然后利用长度和给定的n计算要删除的元素的位置，再遍历到该位置，将它删除。注意如果是第一个的话要特殊处理。

代码：

public ListNode removeNthFromEnd(ListNode head, int n){    ListNode tmp = head;    int length = 0;    while(tmp != null)    {        length++;        tmp = tmp.next;    }    int pos = length - n + 1;    tmp = head;    if(pos == 1)        return head.next;    int i = 1;    while(tmp != null)    {        if(++i == pos)        {            tmp.next = tmp.next.next;            break;        }        tmp = tmp.next;    }    return head;}

结果细节（图）：