Delete all occurrences of a given key in a linked list

Last Updated : 04 Sep, 2024

Given a singly linked list, the task is to delete all occurrences of a given key in it.

Examples:

Input: head: 2 -> 2 -> 1 -> 8 -> 2 -> NULL, key = 2
Output: 1 -> 8 -> NULL

Explanation: All occurrences of the given key = 2, is deleted from the Linked List

Input: head: 1 -> 1 -> 1 -> 4 -> 1 -> NULL, key = 1
Output: 4 -> NULL

Explanation: All occurrences of the given key = 1, is deleted from the Linked List.

Approach:

The Idea is to traverse the list while maintaining a prev pointer to track the previous node. If the current node contains the key, update the next pointer of the prev node to the current node, effectively removing it from the list. This process continues until the end of the list, ensuring all nodes with the specified key are removed.

Below is the implementation of the above idea:

C++

// Iterative C++ program to delete all occurrences of a 
// given key in a linked list
#include <iostream>
using namespace std;

class Node {
public:
    int data;
    Node* next;

    Node(int new_data) {
        data = new_data;
        next = nullptr;
    }
};

// Given the head of a list, delete all occurrences of a 
// given key and return the new head of the list
Node* deleteOccurrences(Node* head, int key) {

    // Initialize pointers to traverse the linked list
    Node *curr = head, *prev = nullptr;

    // Traverse the list to delete all occurrences
    while (curr != nullptr) {

        // If current node's data is equal to key
        if (curr->data == key) {

            // If node to be deleted is head node
            if (prev == nullptr) {
                head = curr->next;
            } 
          
            else {
                prev->next = curr->next;
            }

            // Move to the next node
            curr = curr->next;

        } 
        
        else {
          
            // Move pointers one position ahead
            prev = curr;
            curr = curr->next;
        }
    }

    return head;
}

void printList(Node* curr) {
    while (curr != nullptr) {
        cout << " " << curr->data;
        curr = curr->next;
    }
}

int main() {

    // Create a hard-coded linked list:
    // 2 -> 2 -> 1 -> 8 -> 2 -> NULL
    Node* head = new Node(2);
    head->next = new Node(2);
    head->next->next = new Node(1);
    head->next->next->next = new Node(8);
    head->next->next->next->next = new Node(2);

    int key = 2;
    head = deleteOccurrences(head, key);
    printList(head);
    return 0;
}

// Iterative C program to delete all occurrences of a 
// given key in a linked list
#include <stdio.h>
#include <stdlib.h>

struct Node {
    int data;
    struct Node* next;
};

// Given the head of a list, delete all occurrences of a 
// given key and return the new head of the list
struct Node* deleteOccurrences(struct Node* head, int key) {

    // Initialize pointers to traverse the linked list
    struct Node *curr = head, *prev = NULL;

    // Traverse the list to delete all occurrences
    while (curr != NULL) {

        // If current node's data is equal to key
        if (curr->data == key) {

            // If node to be deleted is head node
            if (prev == NULL) {
                head = curr->next;
            } 
          
            else {
                prev->next = curr->next;
            }

            // Move to the next node
            curr = curr->next;

        } 
      
         else {
           
            // Move pointers one position ahead
            prev = curr;
            curr = curr->next;
        }
    }

    return head;
}

void printList(struct Node* curr) {
    while (curr != NULL) {
        printf(" %d", curr->data);
        curr = curr->next;
    }
}

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:
    // 2 -> 2 -> 1 -> 8 -> 2 -> NULL
    struct Node* head = createNode(2);
    head->next = createNode(2);
    head->next->next = createNode(1);
    head->next->next->next = createNode(8);
    head->next->next->next->next = createNode(2);

    int key = 2;
    head = deleteOccurrences(head, key);
    printList(head);

    return 0;
}

Java

// Java program to delete all occurrences of a given key 
// in a linked list
class Node {
    int data;
    Node next;

    Node(int new_data) {
        data = new_data;
        next = null;
    }
}

// Given the head of a list, delete all occurrences of a 
// given key and return the new head of the list
public class GfG {

    static Node deleteOccurrences(Node head, int key) {
        
        // Initialize pointers to traverse the linked list
        Node curr = head, prev = null;

        // Traverse the list to delete all occurrences
        while (curr != null) {

            // If current node's data is equal to key
            if (curr.data == key) {

                // If node to be deleted is head node
                if (prev == null) {
                    head = curr.next;
                } 
              
                else {
                    prev.next = curr.next;
                }

                // Move to the next node
                curr = curr.next;

            } 
          
             else {
               
                // Move pointers one position ahead
                prev = curr;
                curr = curr.next;
            }
        }

        return head;
    }

    // This function prints the contents of the linked list 
    // starting from the head
    static void printList(Node curr) {
        while (curr != null) {
            System.out.print(" " + curr.data);
            curr = curr.next;
        }
    }

    public static void main(String[] args) {

        // Create a hard-coded linked list:
        // 2 -> 2 -> 1 -> 8 -> 2 -> NULL
        Node head = new Node(2);
        head.next = new Node(2);
        head.next.next = new Node(1);
        head.next.next.next = new Node(8);
        head.next.next.next.next = new Node(2);

        int key = 2;

        head = deleteOccurrences(head, key);
        printList(head);
    }
}

Python

# Python program to delete all occurrences of a given key 
# in a singly linked list
class Node:
    def __init__(self, new_data):
        self.data = new_data
        self.next = None

# Given the head of a list, delete all occurrences of a 
# given key and return the new head of the list
def delete_occurrences(head, key):

    # Initialize pointers to traverse the linked list
    curr = head
    prev = None

    # Traverse the list to delete all occurrences
    while curr is not None:

        # If current node's data is equal to key
        if curr.data == key:

            # If node to be deleted is head node
            if prev is None:
                head = curr.next
            else:
                prev.next = curr.next

            # Move to the next node
            curr = curr.next

        else:
            # Move pointers one position ahead
            prev = curr
            curr = curr.next

    return head

def print_list(curr):
    while curr is not None:
        print(f" {curr.data}", end="")
        curr = curr.next
    print()

if __name__ == "__main__":

    # Create a hard-coded linked list:
    # 2 -> 2 -> 1 -> 8 -> 2 -> None
    head = Node(2)
    head.next = Node(2)
    head.next.next = Node(1)
    head.next.next.next = Node(8)
    head.next.next.next.next = Node(2)

    key = 2
    head = delete_occurrences(head, key)
    print_list(head)

// C# program to delete all occurrences of a given key 
// in a singly linked list
using System;

class Node {
    public int Data;
    public Node next;

    public Node(int newData) {
        Data = newData;
        next = null;
    }
}

class GfG {
  
   // Function to delete all occurrences of a 
   // given key and return the new head of the list
    static Node DeleteOccurrences(Node head, int key) {
        
        // Initialize pointers to traverse the linked list
        Node curr = head;
        Node prev = null;

        // Traverse the list to delete all occurrences
        while (curr != null) {

            // If current node's data is equal to key
            if (curr.Data == key) {

                // If node to be deleted is head node
                if (prev == null) {
                    head = curr.next;
                } 
           
                else {
                    prev.next = curr.next;
                }

                // Move to the next node
                curr = curr.next;

            } 
          
            else {
              
                // Move pointers one position ahead
                prev = curr;
                curr = curr.next;
            }
        }

        return head;
    }

    static void PrintList(Node curr) {
        while (curr != null) {
            Console.Write(" " + curr.Data);
            curr = curr.next;
        }
        Console.WriteLine();
    }

    static void Main() {
         
        // Create a hard-coded linked list:
        // 2 -> 2 -> 1 -> 8 -> 2 -> NULL
        Node head = new Node(2);
        head.next = new Node(2);
        head.next.next = new Node(1);
        head.next.next.next = new Node(8);
        head.next.next.next.next = new Node(2);

        int key = 2;
        head = DeleteOccurrences(head, key);
        PrintList(head);
    }
}

JavaScript

// JavaScript program to delete all occurrences of a given key 
// in a singly linked list
class Node {
    constructor(newData) {
        this.data = newData;
        this.next = null;
    }
}

// Given the head of a list, delete all occurrences of a given 
// key and return the new head of the list
function deleteOccurrences(head, key) {

    // Initialize pointers to traverse the linked list
    let curr = head;
    let prev = null;

    // Traverse the list to delete all occurrences
    while (curr !== null) {

        // If current node's data is equal to key
        if (curr.data === key) {

            // If node to be deleted is head node
            if (prev === null) {
                head = curr.next;
            } 
            
            else {
                prev.next = curr.next;
            }

            // Move to the next node
            curr = curr.next;

        } 
        
        else {
        
            // Move pointers one position ahead
            prev = curr;
            curr = curr.next;
        }
    }

    return head;
}

function printList(curr) {
    while (curr !== null) {
        console.log(" " + curr.data);
        curr = curr.next;
    }
    console.log();
}

// Create a hard-coded linked list:
// 2 -> 2 -> 1 -> 8 -> 2 -> NULL
let head = new Node(2);
head.next = new Node(2);
head.next.next = new Node(1);
head.next.next.next = new Node(8);
head.next.next.next.next = new Node(2);

let key = 2;
head = deleteOccurrences(head, key);
printList(head);

Output

1 8

Time complexity: O(n), where n is the number of nodes in the list.
Auxiliary Space: O(1)

kartik

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