[Leetcode] [Java] Remove Nth Node from End of List

Title:

Given A linked list, remove the nth node from the end of the list and return its head.

For example,

   1->2->3->4->5  N = 2.   1->2->3->5.

Note:
Given n would always be valid.
Try to do the in one pass.

Test Instructions:

Given a linked list, remove the nodes of the nth node of the tail nodes and return the head node after the node is removed.

For example:

   1->2->3->4->5  N = 2.   1->2->3->5.

Note:

The given n is valid.

Try to get the results at once.

Algorithm Analysis:

Use fast and slow pointers. The fast pointer is n steps ahead of the slow pointer. When the fast reaches the end, the slow pointer points at the previous element of the target element.

Use a fast and slow double pointer. The fast pointer is n steps before the slow pointer. When the fast pointer reaches the end of the list, the slow pointer points exactly to the first element of the target element.

AC Code:

public class Solution {public    ListNode removenthfromend (listnode head, int n)     {        if (head = = null)            return null;             ListNode fast = head;        ListNode slow = head;             for (int i=0; i<n; i++)        {            fast = Fast.next;        }             If remove the first node        if (fast = = null)        {            head = Head.next;            return head;        }             while (fast.next! = null)        {            fast = Fast.next;            slow = Slow.next;        }             Slow.next = Slow.next.next;             return head;    }}

Copyright NOTICE: This article is the original article of Bo Master, reprint annotated source

[Leetcode] [Java] Remove Nth Node from End of List