Delete all the nodes from the list which are less than K

Given a Linked List and a key K. The task is to write a program to delete all the nodes from the list that are lesser than the key K.

Examples:

Input :12->15->9->11->5->6
        K = 9
Output : 12 -> 15 -> 9 -> 11

Input : 13->4->16->9->22->45->5->16->6
        K = 10
Output : 13 -> 16 -> 22 -> 45 -> 16

Approach: The approach is similar to deleting all nodes from the list which are greater than the given key.



There are two possible cases:

  1. Head node holds a value less than K: First check for all occurrences at head node which are lesser than ‘K’, delete them and change the head node appropriately.
  2. Some intermediate node holds value less than k: Start traversing from head and check if current node’s value is less than K. If yes then delete that node and move forward in the list.

Below is the implementation of the above approach:

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// C++ program to delete all the nodes from the list
// that are lesser than the specified value K
#include <bits/stdc++.h>
  
using namespace std;
  
// structure of a node
struct Node {
    int data;
    Node* next;
};
  
// function to get a new node
Node* getNode(int data)
{
    Node* newNode = new Node;
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}
  
// function to delete all the nodes from the list
// that are lesser than the specified value K
void deleteLesserNodes(Node** head_ref, int K)
{
    Node *temp = *head_ref, *prev;
  
    // If head node itself holds the value lessser than 'K'
    while (temp != NULL && temp->data < K) {
        *head_ref = temp->next; // Changed head
        free(temp); // free old head
        temp = *head_ref; // Change temp
    }
  
    // Delete occurrences other than head
    while (temp != NULL) {
  
        // Search for the node to be deleted,
        // keep track of the previous node as we
        // need to change 'prev->next'
        while (temp != NULL && temp->data >= K) {
            prev = temp;
            temp = temp->next;
        }
  
        // If required value node was not present
        // in linked list
        if (temp == NULL)
            return;
  
        // Unlink the node from linked list
        prev->next = temp->next;
  
        delete temp; // Free memory
  
        // Update Temp for next iteration of
        // outer loop
        temp = prev->next;
    }
}
  
// function to a print a linked list
void printList(Node* head)
{
    while (head) {
        cout << head->data << " ";
        head = head->next;
    }
}
  
// Driver code
int main()
{
    // Create list: 12->15->9->11->5->6
    Node* head = getNode(12);
    head->next = getNode(15);
    head->next->next = getNode(9);
    head->next->next->next = getNode(11);
    head->next->next->next->next = getNode(5);
    head->next->next->next->next->next = getNode(6);
  
    int K = 9;
  
    cout << "Initial List: ";
    printList(head);
  
    deleteLesserNodes(&head, K);
  
    cout << "\nFinal List: ";
    printList(head);
  
    return 0;
}

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Output:

Initial List: 12 15 9 11 5 6 
Final List: 12 15 9 11

Time Complexity: O(N)



My Personal Notes arrow_drop_up

Strategy Path planning and Destination matters in success No need to worry about in between temporary failures

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