LeetCode 19. Remove Nth Node From End of List（删除链表）

原题网址：https://leetcode.com/problems/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.
方法：设置计数器。

/**
* 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) {
ListNode delete = head;
int pos = -n;
ListNode current = head;
while (current.next != null) {
current = current.next;
pos ++;
if (pos > 0) delete = delete.next;
}
if (pos == -1) return head.next;
delete.next = delete.next.next;
return head;
}
}

另一种实现方式：

/**
* 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) {
ListNode start = new ListNode(0);
start.next = head;
int count = 0;
ListNode current = start;
ListNode remove = start;
while (current.next != null) {
count ++;
current = current.next;
if (count > n) remove = remove.next;
}
remove.next = remove.next.next;
return start.next;
}
}