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Given a singly linked list, write a code to remove nth node from the end of a linked list. In this problem, you can assume the value of n is always valid.

Tóm lược đại ý quan trọng trong bài

• Remove Nth Node from End of a Linked List Java Code
• Remove Nth Node from End of List using One Pass/Traversal Java Code
• Remove Nth Node from List End InterviewBit Solution Java Code
• Related posts:

For example

In this example, the input linked list has four nodes and we have to remove 2nd node from the end.

Input

15 -> 9 -> 8 -> 5 -> NULL , N = 2

After removing second node (node whose value is 8) from the end, the output is

15 -> 9 -> 5 -> NULL

We have discussed the problem statement. Lets discuss multiple approaches and their time and space complexities to solve this problem.

Programming Video Tutorials

Reverse a linked list

Remove Nth Node from End of a Linked List Java Code

In this approach, we are going to traverse a linked list twice to solve this problem.

In the first traversal, we count the number of nodes present in a linked list. Once we know the count of elements in a list, the next step is to remove nth node from the end.

The time complexity of this approach is O(n) and its space complexity is O(1).

Lets write its java code.

Java1234567891011//Structure of node classpublic class Node int data;Node next;public Node(int data) this.data = data; this.next = null; Java1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283848586878889//Remove node from end using two passpublic class RemoveNodeFromEnd public Node insertAtHead(Node head, int data) if (head == null) head = new Node(data); else // Create a new node Node temp = new Node(data); // New node points to head temp.next = head; // Head points to a new node head = temp; return head;public Node removeNodeFromEnd(Node head, int n) Node temp = head;int len = 0;//Count number of nodes while(temp != null) temp = temp.next; len++;int pos = len–n+1;//If only one node is there in linked listif(pos == 1) return head.next; temp = head; len = 0;while(temp != null) len++; if(pos–1 == len) temp.next = temp.next.next; temp = temp.next;return head;public void print(Node head) Node temp = head;while (temp != null) System.out.print(temp.data + ” -> “);temp = temp.next; System.out.print(“null”); System.out.println(” “); public static void main(String[] args) Node head = null; RemoveNodeFromEnd ll = new RemoveNodeFromEnd(); head = ll.insertAtHead(head, 5); head = ll.insertAtHead(head, 8); head = ll.insertAtHead(head, 9); head = ll.insertAtHead(head, 15); System.out.println(” Linked List “); ll.print(head); Node result = ll.removeNodeFromEnd(head, 2); System.out.println(” After removing element “); ll.print(result);

Programming questions on linked list

Programming questions on binary tree

Remove Nth Node from End of List using One Pass/Traversal Java Code

We can solve this problem in a single traversal/pass using two pointers (slow and fast).

To solve this problem, First, we need to move fast pointer by n steps. After that move fast and slow pointer by one step until fast pointer does not point to null.

When the fast pointer reaches at the end the slow pointer is exactly nth node from end.

The time complexity of removing nth node from end is O(n) and its space complexity is O(1).

Remove duplicates from a sorted linked list

Java123456789101112131415161718192021222324252627282930//Single traversal java codepublic Node removeNodeFromEndSinglePass(Node head, int n) //Declare two pointersNode slow = head;Node fast = head;//Move fast pointer by n stepsfor(int i = 1; i <= n; i++) fast = fast.next;/* Special handling, when only one node is present ina linked list*/if(fast == null) head = head.next; return head; while(fast.next != null) fast = fast.next; slow = slow.next; slow.next = slow.next.next;return head;

Remove Nth Node from List End InterviewBit Solution Java Code

In this problem also we have to remove the nth node from the end of list.

Here, If the value of n is greater than the size of the list, then we have to remove the first node of the list.

Java12345678910111213141516171819202122232425262728293031323334353637//Java Codepublic class Solution public ListNode removeNthFromEnd(ListNode A, int B) int count = 0;//Declare two pointers slow and fastListNode slow = A;ListNode fast = A;while(count++ <= B && fast != null) fast = fast.next;/* If the value of B is greater the no. of elements present in a list then remove the first element of a linked list.*/if(count < B)return A.next;if(fast == null)return slow.next;while(fast != null)fast = fast.next;slow = slow.next;slow.next = slow.next.next;return A;ShareTweetShare0 Shares

Related posts:

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C Program to Reverse a Linked List using Recursion

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Odd Even Linked List

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