19.Remove Nth Node From End of List（7.54%）

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.

题目解释：给你一个链表，告诉你倒数第n个节点要移除，你要做的就是去掉该链表的倒数第n个节点，然后将该链表返回。题目说，输入的n肯定会有效的，所以不用考虑n无效的情况。

用最快的速度，遍历一次搞定。

/**
* 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) {
int index = 0;
ListNode removedList = head;
ListNode scanList = head;
while (scanList.next != null) {
// 假设刚好只有n个链表
if (index >= n) {
removedList = removedList.next;
}
scanList = scanList.next;
index++;
}

index++;
// 移除第一个
if (index == n) {
return head.next;

}

removedList.next = removedList.next.next;
return head;
}
}