Remove Nth Node From End of List LeetCode Solution | Easy Approach

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

Remove Nth Node From End of List Solutions

✅Time: O(n)
✅Space: O(1)

C++

class Solution {
 public:
  ListNode* removeNthFromEnd(ListNode* head, int n) {
    auto slow = head;
    auto fast = head;

    while (n--)
      fast = fast->next;
    if (!fast)
      return head->next;

    while (fast->next) {
      slow = slow->next;
      fast = fast->next;
    }
    slow->next = slow->next->next;

    return head;
  }
};

Java

class Solution {
  public String intToRoman(int num) {
    final int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
class Solution {
  public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode slow = head;
    ListNode fast = head;

    while (n-- > 0)
      fast = fast.next;
    if (fast == null)
      return head.next;

    while (fast.next != null) {
      slow = slow.next;
      fast = fast.next;
    }
    slow.next = slow.next.next;

    return head;
  }
}

Python

class Solution:
  def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
    slow = head
    fast = head

    for _ in range(n):
      fast = fast.next
    if not fast:
      return head.next

    while fast.next:
      slow = slow.next
      fast = fast.next
    slow.next = slow.next.next

    return head

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