Losing head/tail while rewiring
Wrong move: Pointer updates overwrite references before they are saved.
Usually fails on: List becomes disconnected mid-operation.
Fix: Store next pointers first and use a dummy head for safer joins.
Move from brute-force thinking to an efficient approach using linked list strategy.
There is a singly-linked list head and we want to delete a node node in it.
You are given the node to be deleted node. You will not be given access to the first node of head.
All the values of the linked list are unique, and it is guaranteed that the given node node is not the last node in the linked list.
Delete the given node. Note that by deleting the node, we do not mean removing it from memory. We mean:
node should be in the same order.node should be in the same order.Custom testing:
head and the node to be given node. node should not be the last node of the list and should be an actual node in the list.Example 1:
Input: head = [4,5,1,9], node = 5 Output: [4,1,9] Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.
Example 2:
Input: head = [4,5,1,9], node = 1 Output: [4,5,9] Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.
Constraints:
[2, 1000].-1000 <= Node.val <= 1000node to be deleted is in the list and is not a tail node.Problem summary: There is a singly-linked list head and we want to delete a node node in it. You are given the node to be deleted node. You will not be given access to the first node of head. All the values of the linked list are unique, and it is guaranteed that the given node node is not the last node in the linked list. Delete the given node. Note that by deleting the node, we do not mean removing it from memory. We mean: The value of the given node should not exist in the linked list. The number of nodes in the linked list should decrease by one. All the values before node should be in the same order. All the values after node should be in the same order. Custom testing: For the input, you should provide the entire linked list head and the node to be given node. node should not be the last node of the list and should be an actual node in the list. We will build the linked list and pass the node to
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Linked List
[4,5,1,9] 5
[4,5,1,9] 1
remove-linked-list-elements)remove-nodes-from-linked-list)delete-nodes-from-linked-list-present-in-array)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #237: Delete Node in a Linked List
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public void deleteNode(ListNode node) {
node.val = node.next.val;
node.next = node.next.next;
}
}
// Accepted solution for LeetCode #237: Delete Node in a Linked List
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func deleteNode(node *ListNode) {
node.Val = node.Next.Val
node.Next = node.Next.Next
}
# Accepted solution for LeetCode #237: Delete Node in a Linked List
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def deleteNode(self, node):
"""
:type node: ListNode
:rtype: void Do not return anything, modify node in-place instead.
"""
node.val = node.next.val
node.next = node.next.next
// Accepted solution for LeetCode #237: Delete Node in a Linked List
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #237: Delete Node in a Linked List
// /**
// * Definition for singly-linked list.
// * public class ListNode {
// * int val;
// * ListNode next;
// * ListNode(int x) { val = x; }
// * }
// */
// class Solution {
// public void deleteNode(ListNode node) {
// node.val = node.next.val;
// node.next = node.next.next;
// }
// }
// Accepted solution for LeetCode #237: Delete Node in a Linked List
/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
/**
Do not return anything, modify it in-place instead.
*/
function deleteNode(node: ListNode | null): void {
node.val = node.next.val;
node.next = node.next.next;
}
Use this to step through a reusable interview workflow for this problem.
Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.
Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.
Review these before coding to avoid predictable interview regressions.
Wrong move: Pointer updates overwrite references before they are saved.
Usually fails on: List becomes disconnected mid-operation.
Fix: Store next pointers first and use a dummy head for safer joins.