LeetCode – Delete Node in a Linked List (Java)

 

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.

Java Solution

public void deleteNode(ListNode node) {
    node.val = node.next.val;
    node.next = node.next.next;
}

Is this problem too easy? Or I'm doing it wrong.

Related posts:

  1. LeetCode – Remove Linked List Elements (Java)
  2. LeetCode – Linked List Random Node (Java)
  3. LeetCode – Remove Nth Node From End of List (Java)
  4. LeetCode – Reverse Linked List (Java)
Category >> Algorithms >> Interview  
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  1. CRH on 2015-12-24

    This is easy, BUT needs edge cases handling in code/talking to your interviewer. Did you wonder how this will work if the node is the last in the list? 🙂

  2. Aditya Vutukuri on 2016-2-27

    You need to take care if this is a head and tail.

    public void deleteNode(ListNode n) {

    node.next = node.next.next;
    }

  3. chintan thakker on 2016-9-12

    Your code is correct for somewhat different Question i.e remove elements if access to head is given but somewhat duplicated. There is no need to have seperate condition for last node, Change while(curr.next != null) to while(curr!=null). It works perfectly. Also check if Head is null(i.e if the list is empty otherwise curr = head.next; will throw a null pointer exception)

    1st Method:-

    public void deleteNode(ListNode n) {

    if(head ==null){
    System.out.println("List is Empty");
    }
    if(head.val == n){ //Check if the first node is the one to be deleted
    head=head.next;
    }
    if(head == null){ // Check if there was only one element on Original list
    return;
    }

    ListNode prev = head;
    ListNode curr = head.next;

    while (curr != null){
    if(curr.val == n.val){
    prev.next = curr.next;
    }else{
    prev = prev.next;
    curr = curr.next // no null pointer exception will be shown even if curr.next is null
    }
    }

    }

    2nd method:-


    public void deleteNode(ListNode n) {

    if(head ==null){
    System.out.println("List is Empty");
    }
    if(head.val == n){
    head=head.next;
    }
    if(head == null){ // Check if there was only one element in Original list
    return;
    }
    ListNode curr = head;
    while (curr.next != null) {
    if (curr.next.val == n) {
    curr.next= curr.next.next;
    } else {
    curr = curr.next;
    }
    }
    }

  4. jp19online on 2017-2-21

    The solution is technically incorrect if the node has other meta data. In reality – it may be important to delete that exact node’s memory and not the next one’s

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