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）：