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; } |
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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 |
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Python
# Python code to implement binary search # on Singly Linked List # Link list node class Node: def __init__(self, data): self.data = data self.next = None self.prev = None def newNode(x): temp = Node(0) temp.data = x temp.next = None return temp # function to find out middle element def middle(start, last): if (start == None): return None slow = start 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 def binarySearch(head,value): start = head last = None while True : # Find middle mid = middle(start, last) # If middle is empty if (mid == None): return None # If value is present at middle if (mid . data == value): return mid # If value is more than mid elif (mid . data < value): start = mid . next # If the value is less than mid. else: last = mid if not (last == None or last != start): break # value not present return None # Driver Code 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) value = 7if (binarySearch(head, value) == None): print("Value not present\n") else: print("Present") # This code is contributed by Arnab Kundu |
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C#
// C# code to implement binary search // on Singly Linked List using System; // Node Class public class Node { public int data; public Node next; // Constructor to create a new node public 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) { Console.WriteLine("Value not present"); } else { Console.WriteLine("Present"); } } } // This code is contributed by Arnab Kundu |
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Output:
Present
Time Complexity : O(n)
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