**Remove Nth Node From End of List LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Problems C++**

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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 `n`

node from the end of the list and return its head.^{th}

**Example 1:**

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

**Example 2:**

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

**Example 3:**

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