Linked Structures

Delete by Value

Find the node before the target value and rewire its next pointer so the target node leaves the chain.

Algorithm

The replay labels nodes by value, such as node(20), and never exposes object identity or memory addresses. This Swift DSA implementation uses the same small chain as the rest of the DSA track.

Basic Implementation

basic.swift 
final class Node {
    let value: Int
    var next: Node?
    init(_ value: Int, _ next: Node? = nil) {
        self.value = value
        self.next = next
    }
}

func render(_ head: Node?) -> String {
    var parts: [String] = []
    var cursor = head
    while cursor != nil {
        parts.append(String(cursor!.value))
        cursor = cursor!.next
    }
    return parts.joined(separator: " -> ") + " -> null"
}

var head: Node? = Node(10, Node(20, Node(30, Node(40))))
let target = 30
var cursor = head
while cursor?.next != nil {
    if cursor!.next!.value == target {
        cursor!.next = cursor!.next!.next
        break
    }
    cursor = cursor!.next
}
print(render(head))

Complexity

  • Time: O(n)
  • Space: O(1)

Implementation notes

  • Keep the explicit node and pointer/reference operations; array shortcuts hide the linked-list state this lesson is meant to replay.
  • The final output prints the chain in a deterministic a -> b -> null form for cross-language comparison.
predecessor Deletion needs the node before the one being removed.
rewiring The predecessor skips the target and points at the target's next node.