Remove Nth Node From End of List

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

 

Example 1:

Input: head = [1,2,3,4,5], n = 2

Output: [1,2,3,5]

Example 2:

Input: head = [1], n = 1

Output: []

Example 3:

Input: head = [1,2], n = 1

Output: [1]

 

Constraints:

The number of nodes in the list is sz.

1 <= sz <= 30

0 <= Node.val <= 100

1 <= n <= sz

 

Follow up: Could you do this in one pass?

🔷 Step-by-Step Logic:

1. First traverse the list once to count total nodes.

2. Then calculate the position to delete from the start: pos = total - n

3. Traverse again up to that pos.

4. Use two pointers: prev and current.

5. When pos == 0, skip the node by: prev.next = current.next;

/**

 * Definition for singly-linked list.

 * public class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode() {}

 *     ListNode(int val) { this.val = val; }

 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }

 * }

 */

class Solution {

  public ListNode removeNthFromEnd(ListNode head, int n) {

    int count = 0;

    ListNode temp = head;

    // 1. Count total nodes

    while (temp != null) {

        count++;

        temp = temp.next;

    }

    // 2. If we need to remove the head node

    if (count == n) {

        return head.next;

    }

    // 3. Move to (count - n)th node

    int pos = count - n;

    ListNode prev = null;

    ListNode current = head;

    while (pos > 0) {

        prev = current;

        current = current.next;

        pos--;

    }

    // 4. Remove the target node

    prev.next = current.next;

    return head;

}

}