Binary Search on Singly Linked List

Given a singly linked list and a key, find key using binary search approach.
To perform a Binary search based on Divide and Conquer Algorithm, determination of the middle element is important. Binary Search is usually fast and efficient for arrays because accessing the middle index between two given indices is easy and fast(Time Complexity O(1)). But memory allocation for the singly linked list is dynamic and non-contiguous, which makes finding the middle element difficult. One approach could be of using skip list, one could be traversing the linked list using one pointer.

Prerequisite : Finding middle of a linked list.

Note: The approach and implementation provided below are to show how Binary Search can be implemented on a linked list. The implementation takes O(n) time.

Approach :

Here, start node(set to Head of list), and the last node(set to NULL initially) are given.
Middle is calculated using two pointers approach.
If middle’s data matches the required value of search, return it.
Else if middle’s data < value, move to upper half(setting start to middle's next).
Else go to lower half(setting last to middle).
The condition to come out is, either element found or entire list is traversed. When entire list is traversed, last points to start i.e. last -> next == start.

In main function, function InsertAtHead inserts value at the beginning of linked list. Inserting such values(for sake of simplicity) so that the list created is sorted.

Examples :

Input : Enter value to search : 7
Output : Found

Input : Enter value to search : 12
Output : Not Found

C++

// CPP code to implement binary search 
// on Singly Linked List 
#include<stdio.h> 
#include<stdlib.h> 
  
struct Node 
{ 
    int data; 
    struct Node* next; 
}; 
  
Node *newNode(int x) 
{ 
    struct Node* temp = new Node; 
    temp->data = x; 
    temp->next = NULL; 
    return temp; 
} 
  
// function to find out middle element 
struct Node* middle(Node* start, Node* last) 
{ 
    if (start == NULL) 
        return NULL; 
  
    struct Node* slow = start; 
    struct Node* fast = start -> next; 
  
    while (fast != last) 
    { 
        fast = fast -> next; 
        if (fast != last) 
        { 
            slow = slow -> next; 
            fast = fast -> next; 
        } 
    } 
  
    return slow; 
} 
  
// Function for implementing the Binary 
// Search on linked list 
struct Node* binarySearch(Node *head, int value) 
{ 
    struct Node* start = head; 
    struct Node* last = NULL; 
  
    do
    { 
        // Find middle 
        Node* mid = middle(start, last); 
  
        // If middle is empty 
        if (mid == NULL) 
            return NULL; 
  
        // If value is present at middle 
        if (mid -> data == value) 
            return mid; 
  
        // If value is more than mid 
        else if (mid -> data < value) 
            start = mid -> next; 
  
        // If the value is less than mid. 
        else
            last = mid; 
  
    } while (last == NULL || 
             last != start); 
  
    // value not present 
    return NULL; 
} 
  
// Driver Code 
int main() 
{ 
    Node *head = newNode(1); 
    head->next = newNode(4); 
    head->next->next = newNode(7); 
    head->next->next->next = newNode(8); 
    head->next->next->next->next = newNode(9); 
    head->next->next->next->next->next = newNode(10); 
    int value = 7; 
    if (binarySearch(head, value) == NULL) 
        printf("Value not present\n"); 
    else
        printf("Present"); 
    return 0; 
} 

Java

// Java code to implement binary search  
// on Singly Linked List  
  
// Node Class 
class Node 
{ 
    int data; 
    Node next; 
  
    // Constructor to create a new node 
    Node(int d) 
    { 
        data = d; 
        next = null; 
    } 
} 
  
class BinarySearch 
{ 
    // function to insert a node at the beginning 
    // of the Singaly Linked List 
    static Node push(Node head, int data) 
    { 
        Node newNode = new Node(data); 
        newNode.next = head; 
        head = newNode; 
        return head; 
    } 
  
    // Function to find middle element 
    // using Fast and Slow pointers 
    static Node middleNode(Node start, Node last)  
    { 
        if (start == null) 
            return null; 
  
        Node slow = start; 
        Node fast = start.next; 
  
        while (fast != last) 
        { 
            fast = fast.next; 
            if (fast != last)  
            { 
                slow = slow.next; 
                fast = fast.next; 
            } 
        } 
        return slow; 
    } 
  
    // function to insert a node at the beginning 
    // of the Singly Linked List 
    static Node binarySearch(Node head, int value)  
    { 
        Node start = head; 
        Node last = null; 
  
        do
        { 
            // Find Middle 
            Node mid = middleNode(start, last); 
  
            // If middle is empty 
            if (mid == null) 
                return null; 
  
            // If value is present at middle 
            if (mid.data == value) 
                return mid; 
  
            // If value is less than mid 
            else if (mid.data > value)  
            { 
                start = mid.next; 
            } 
  
            // If the value is more than mid. 
            else
                last = mid; 
        } while (last == null || last != start); 
  
        // value not present 
        return null; 
    } 
  
    // Driver Code 
    public static void main(String[] args)  
    { 
        Node head = null; 
  
        // Using push() function to 
        // convert singly linked list 
        // 10 -> 9 -> 8 -> 7 -> 4 -> 1 
        head = push(head, 1); 
        head = push(head, 4); 
        head = push(head, 7); 
        head = push(head, 8); 
        head = push(head, 9); 
        head = push(head, 10); 
        int value = 7; 
  
        if (binarySearch(head, value) == null) 
        { 
            System.out.println("Value not present"); 
        }  
        else
        { 
            System.out.println("Present"); 
        } 
    } 
} 
  
// This code is contributed by Vivekkumar Singh 

Output:

Present

Time Complexity : O(n)

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