Given a linked list, remove the n-th node from the end of list and return its head.
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.
Follow up:
Could you do this in one pass?
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { if(head == null) return null; ListNode fast = head; while(n-- > 0) { fast = fast.next; } if(fast == null) return head.next; ListNode slow = head; while(fast.next != null) { slow = slow.next; fast = fast.next; } slow.next = slow.next.next; return head; } }