Insert node at middle of linked list in swift

Suppose we are inserted the following (1, 2, 3, 4, 5, 6, 7) node in a sequence. In this post are implement second approach.

insert linked list node in middle position
import Foundation
// Swift 4 Program for
// Insert linked list element at middle position

// Linked list node
class LinkNode
{
    var data: Int;
    var next: LinkNode? ;
    init(_ data: Int)
    {
        self.data = data;
        self.next = nil;
    }
}
class LinkedList
{
    var head: LinkNode? ;
    // Class constructors
    init()
    {
        self.head = nil;
    }
    // Insert node in middle position
    func insert(_ value: Int)
    {
        // Create a new node
        let node: LinkNode? = LinkNode(value);
        if (self.head == nil)
        {
            // First node
            self.head = node;
        }
        else
        {
            var temp: LinkNode? = self.head;
            var middle: LinkNode? = self.head;
            // Find the middle node
            while (temp!.next  != nil && 
              temp!.next!.next  != nil)
            {
                temp = temp!.next!.next;
                middle = middle!.next;
            }
            // add node
            node!.next = middle!.next;
            middle!.next = node;
        }
    }
    // Display linked list element
    func display()
    {
        if (self.head == nil)
        {
            return;
        }
        var temp: LinkNode? = self.head;
        // iterating linked list elements
        while (temp  != nil)
        {
            print(temp!.data, terminator: " → ");
            // Visit to next node
            temp = temp!.next;
        }
        print("NULL");
    }
    static func main(_ args: [String])
    {
        let sll: LinkedList? = LinkedList();
        // Add node
        sll!.insert(1);
        sll!.insert(2);
        sll!.insert(3);
        sll!.insert(4);
        sll!.insert(5);
        sll!.insert(6);
        sll!.insert(7);
        // 1 → 3 → 5 → 7 → 6 → 4 → 2 → NULL
        sll!.display();
    }
}
LinkedList.main([String]());
 1 → 3 → 5 → 7 → 6 → 4 → 2 → NULL



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