# LeetCode-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

### 方法一

##### 算法分析
• 要删除的节点在`L-n+1`位置（`L为链表长度`）。
• 找到第`L-n`个节点，指向`L-n+2`个节点。
##### 答案
``````/**
* 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 dummy = new ListNode(0);
ListNode current = head;
int length = 0;//链表长度
while (current != null) {
length ++;//寻找链表长度
current = current.next;
}
length = length - n;//到要被删除位置的链表节点的前一个
current = dummy;
while (length > 0) {
length --;
current = current.next;
}
current.next = current.next.next;//将被删除的节点前一个节点指向被删除节点的下一个节点
return dummy.next;
}
}
``````

### 方法二

##### 算法分析
• 移动first节点到与second相差n个节点的位置
• 同时移动first、second直到first为null
• 此时second的下一个节点是为要删除的节点
##### 答案
``````/**
* 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 dummy = new ListNode(0);
ListNode first = dummy;
ListNode second = dummy;

for (int i = 1; i <= n + 1; i++) {
first = first.next;//将first移到与second相差n个节点
}
while (first != null) {//first和second同时移动，直到first到最后一个节点的下一个节点
first = first.next;
second = second.next;
}

second.next = second.next.next;
return dummy.next;
}
}
``````

LeetCode