19. Remove Nth Node From End of List LeetCode Solution
In this guide we will provide 19. Remove Nth Node From End of List LeetCode Solution with best time and space complexity. The solution to Remove Nth Node From End of List problem is provided in various programming languages like C++, Java and python. This will be helpful for you if you are preparing for placements, hackathon, interviews or practice purposes. The solutions provided here are very easy to follow and with detailed explanations.
Problem Statement of 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?
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
19. Remove Nth Node From End of List LeetCode Solution in C++
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* slow = head;
ListNode* fast = head;
while (n-- > 0)
fast = fast->next;
if (fast == nullptr)
return head->next;
while (fast->next != nullptr) {
slow = slow->next;
fast = fast->next;
}
slow->next = slow->next->next;
return head;
}
}; /* code provided by PROGIEZ */
19. Remove Nth Node From End of List LeetCode Solution in Java
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;
}
} // code provided by PROGIEZ
19. Remove Nth Node From End of List LeetCode Solution in 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 #code by PROGIEZ
