leetcode-19-remove-nth-node-from-end-of-list.md

Description

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

1
2
3

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;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
       if(head == null){
            return head;
        }
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        ListNode slow = head;
        ListNode fast = head;
        for(int i = 0; i <n; i++){
            fast = fast.next;
        }
        while(fast.next != null){
            fast = fast.next;
            slow = slow.next;
        }
        ListNode temp = slow.next;
        slow.next = slow.next.next;
        return dummy.next;
    }
}