[LeetCode] Remove Nth Node From End of List

Problem:

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.

Solution:O(n)

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(fast == null){

        return head.next;

        }

        while(fast.next != null){

            fast = fast.next;

            slow = slow.next;

        }

        slow.next = slow.next.next;

        return head;

    }

}

Key concept: Using two pointer, faster n step ahead slower.