Binary Search on Singly Linked List Last Updated : 04 Sep, 2024 Comments Improve Suggest changes Like Article Like Report Given a sorted singly linked list and a key, the task is to find the key in the Linked List using Binary Search. Examples: Input: head = 1->4->7->8->9->10, key = 7Output: Present Input: LinkedList = 1->4->7->8->9->10, key = 12Output: Value Not PresentNote that Binary Search does not work efficiently for linked lists. The purpose of this article is to show the same. It is recommended to use simple linear search. If we wish to achieve better than linear time, Skip List is recommended for fast search.Find the middle element of the Linked List.Compare the middle element with the key. if the key is found in the middle element, return true.If the key is not found in the middle element, choose which half will be used as the next search space.If the key is smaller than the middle node, the left half is used for the next search.If the key is greater than the middle node, then the right half is used for the next search.This process is continued until the key is found or the total Linked List is exhausted.Below is the implementation of the above approach: C++ // C++ code to implement binary search // on Singly Linked List #include <bits/stdc++.h> using namespace std; class Node { public: int data; Node *next; Node(int x) { data = x; next = NULL; } }; // function to find out middle element Node *middle(Node *start, Node *last) { if (start == NULL) { return NULL; } if (start == last) return start; Node *slow = start; Node *fast = start->next; while (fast != last) { fast = fast->next; slow = slow->next; if (fast != last) { fast = fast->next; } } return slow; } // Function for implementing the Binary // Search on linked list bool binarySearch(Node *head, int value) { Node *start = head; Node *last = NULL; while (true) { // Find middle Node *mid = middle(start, last); // If middle is empty if (mid == NULL) { return false; } // If value is present at middle if (mid->data == value) return true; // If start and last node are overlapping else if (start == last) break; // If value is more than mid else if (mid->data < value) { start = mid->next; } // If the value is less than mid. else if (mid->data > value) last = mid; } // value not present return false; } int main() { // Create a hard-coded linked list: // 1 -> 4 -> 7 -> 8 -> 9 -> 10 Node *head = new Node(1); head->next = new Node(4); head->next->next = new Node(7); head->next->next->next = new Node(8); head->next->next->next->next = new Node(9); head->next->next->next->next->next = new Node(10); int value = 7; if (binarySearch(head, value)) cout << "Present"; else cout << "Value not present\n"; return 0; } C // C code to implement binary search // on Singly Linked List #include <stdio.h> #include <stdlib.h> struct Node { int data; struct Node *next; }; // function to find out middle element struct Node *middle(struct Node *start, struct Node *last) { if (start == NULL) { return NULL; } if (start == last) return start; struct Node *slow = start; struct Node *fast = start->next; while (fast != last) { fast = fast->next; slow = slow->next; if (fast != last) { fast = fast->next; } } return slow; } // Function for implementing the Binary // Search on linked list int binarySearch(struct Node *head, int value) { struct Node *start = head; struct Node *last = NULL; while (1) { // Find middle struct Node *mid = middle(start, last); // If middle is empty if (mid == NULL) { return 0; } // If value is present at middle if (mid->data == value) return 1; // If start and last node are overlapping else if (start == last) break; // If value is more than mid else if (mid->data < value) { start = mid->next; } // If the value is less than mid. else if (mid->data > value) last = mid; } // value not present return 0; } struct Node *createNode(int new_data) { struct Node *new_node = (struct Node *)malloc(sizeof(struct Node)); new_node->data = new_data; new_node->next = NULL; return new_node; } int main() { // Create a hard-coded linked list: // 1 -> 4 -> 7 -> 8 -> 9 -> 10 struct Node *head = createNode(1); head->next = createNode(4); head->next->next = createNode(7); head->next->next->next = createNode(8); head->next->next->next->next = createNode(9); head->next->next->next->next->next = createNode(10); int value = 7; if (binarySearch(head, value)) printf("Present\n"); else printf("Value not present\n"); return 0; } Java // Java code to implement binary search // on Singly Linked List class Node { int data; Node next; Node(int new_data) { data = new_data; next = null; } } public class GfG { // function to find out middle element static Node middle(Node start, Node last) { if (start == null) { return null; } if (start == last) return start; Node slow = start; Node fast = start.next; while (fast != last) { fast = fast.next; slow = slow.next; if (fast != last) { fast = fast.next; } } return slow; } // Function for implementing the Binary // Search on linked list static boolean binarySearch(Node head, int value) { Node start = head; Node last = null; while (true) { // Find middle Node mid = middle(start, last); // If middle is empty if (mid == null) { return false; } // If value is present at middle if (mid.data == value) return true; // If start and last node are overlapping else if (start == last) break; // If value is more than mid else if (mid.data < value) { start = mid.next; } // If the value is less than mid. else if (mid.data > value) last = mid; } // value not present return false; } public static void main(String[] args) { // Create a hard-coded linked list: // 1 -> 4 -> 7 -> 8 -> 9 -> 10 Node head = new Node(1); head.next = new Node(4); head.next.next = new Node(7); head.next.next.next = new Node(8); head.next.next.next.next = new Node(9); head.next.next.next.next.next = new Node(10); int value = 7; if (binarySearch(head, value)) System.out.println("Present"); else System.out.println("Value not present"); } } Python # Python code to implement binary search # on Singly Linked List class Node: def __init__(self, new_data): self.data = new_data self.next = None # function to find out middle element def middle(start, last): if start is None: return None if start == last: return start slow = start fast = start.next while fast != last: fast = fast.next slow = slow.next if fast != last: fast = fast.next return slow # Function for implementing the Binary # Search on linked list def binary_search(head, value): start = head last = None while True: # Find middle mid = middle(start, last) # If middle is empty if mid is None: return False # If value is present at middle if mid.data == value: return True # If start and last node are overlapping elif start == last: break # If value is more than mid elif mid.data < value: start = mid.next # If the value is less than mid. elif mid.data > value: last = mid # value not present return False if __name__ == "__main__": # Create a hard-coded linked list: # 1 -> 4 -> 7 -> 8 -> 9 -> 10 head = Node(1) head.next = Node(4) head.next.next = Node(7) head.next.next.next = Node(8) head.next.next.next.next = Node(9) head.next.next.next.next.next = Node(10) value = 7 if binary_search(head, value): print("Present") else: print("Value not present") C# // C# code to implement binary search // on Singly Linked List using System; class Node { public int Data; public Node next; public Node(int x) { Data = x; next = null; } } class GfG { // function to find out middle element static Node Middle(Node start, Node last) { if (start == null) { return null; } if (start == last) { return start; } Node slow = start; Node fast = start.next; while (fast != last) { fast = fast.next; slow = slow.next; if (fast != last) { fast = fast.next; } } return slow; } // Function for implementing the Binary // Search on linked list static bool BinarySearch(Node head, int value) { Node start = head; Node last = null; while (true) { // Find middle Node mid = Middle(start, last); // If middle is empty if (mid == null) { return false; } // If value is present at middle if (mid.Data == value) { return true; } // If start and last node are overlapping else if (start == last) { break; } // If value is more than mid else if (mid.Data < value) { start = mid.next; } // If the value is less than mid. else if (mid.Data > value) { last = mid; } } // value not present return false; } static void Main() { // Create a hard-coded linked list: // 1 -> 4 -> 7 -> 8 -> 9 -> 10 Node head = new Node(1); head.next = new Node(4); head.next.next = new Node(7); head.next.next.next = new Node(8); head.next.next.next.next = new Node(9); head.next.next.next.next.next = new Node(10); int value = 7; if (BinarySearch(head, value)) { Console.WriteLine("Present"); } else { Console.WriteLine("Value not present"); } } } JavaScript // JavaScript code to implement binary search // on Singly Linked List class Node { constructor(newData) { this.data = newData; this.next = null; } } // function to find out middle element function middle(start, last) { if (start === null) { return null; } if (start === last) { return start; } let slow = start; let fast = start.next; while (fast !== last) { fast = fast.next; slow = slow.next; if (fast !== last) { fast = fast.next; } } return slow; } // Function for implementing the Binary // Search on linked list function binarySearch(head, value) { let start = head; let last = null; while (true) { // Find middle let mid = middle(start, last); // If middle is empty if (mid === null) { return false; } // If value is present at middle if (mid.data === value) { return true; } // If start and last node are overlapping else if (start === last) { break; } // If value is more than mid else if (mid.data < value) { start = mid.next; } // If the value is less than mid. else if (mid.data > value) { last = mid; } } // value not present return false; } // Create a hard-coded linked list: // 1 -> 4 -> 7 -> 8 -> 9 -> 10 let head = new Node(1); head.next = new Node(4); head.next.next = new Node(7); head.next.next.next = new Node(8); head.next.next.next.next = new Node(9); head.next.next.next.next.next = new Node(10); let value = 7; if (binarySearch(head, value)) { console.log("Present"); } else { console.log("Value not present"); } OutputPresentTime complexity: O(n) , where n is the number of nodes in linked list.Auxilliary Space: O(1) Comment More infoAdvertise with us Next Article Binary Search on Singly Linked List K KunalDhyani Follow Improve Article Tags : Linked List Algorithms Dynamic Programming Divide and Conquer Searching Competitive Programming DSA Binary Search +4 More Practice Tags : AlgorithmsBinary SearchDivide and ConquerDynamic ProgrammingLinked ListSearching +2 More Similar Reads Binary Search Algorithm - Iterative and Recursive Implementation Binary Search Algorithm is a searching algorithm used in a sorted array by repeatedly dividing the search interval in half. 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