Remove Nth Node From End of List LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Remove Nth Node From End of List in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Remove Nth Node From End of List– LeetCode Problem

Remove Nth Node From End of List– LeetCode Problem

Problem:

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– LeetCode Solutions

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;
  }
};

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;
  }
}

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