Priority Queue using Linked List
Last Updated : 11 Feb, 2025
Implement Priority Queue using Linked Lists. The Linked List should be so created so that the highest priority ( priority is assigned from 0 to n-1 where n is the number of elements, where 0 means the highest priority and n-1 being the least ) element is always at the head of the list. The list is arranged in descending order of elements based on their priority. The Priority Queue should have the below functions
- push(): This function is used to insert a new data into the queue.
- pop(): This function removes the element with the highest priority from the queue.
- peek() / top(): This function is used to get the highest priority element in the queue without removing it from the queue.
Examples:
Input: Values = 4 (priority = 1), 5 (priority = 2), 6 (priority = 3), 7 (priority = 0)
Output: 7 4 5 6
Explanation: Values get peeked and popped according to their priority in descending order.
This approach manages a priority queue using a linked list. The PUSH operation inserts nodes in order of priority, ensuring the highest priority node is always at the head. The POP operation removes the highest priority node, while PEEK returns the data of the highest priority node without removing it.
Follow these steps to solve the problem
PUSH(HEAD, DATA, PRIORITY):
- Step 1: Create new node with DATA and PRIORITY
- Step 2: Check if HEAD has lower priority. If true follow Steps 3-4 and end. Else goto Step 5.
- Step 3: NEW -> NEXT = HEAD
- Step 4: HEAD = NEW
- Step 5: Set TEMP to head of the list
- Step 6: While TEMP -> NEXT != NULL and TEMP -> NEXT -> PRIORITY > PRIORITY
- Step 7: TEMP = TEMP -> NEXT
[END OF LOOP] - Step 8: NEW -> NEXT = TEMP -> NEXT
- Step 9: TEMP -> NEXT = NEW
- Step 10: End
POP(HEAD):
- Step 1: Set the head of the list to the next node in the list. HEAD = HEAD -> NEXT.
- Step 2: Free the node at the head of the list
- Step 3: End
PEEK(HEAD):
- Step 1: Return HEAD -> DATA
- Step 2: End
C++ // C++ code to implement Priority Queue // using Linked List #include <bits/stdc++.h> using namespace std; struct Node { int data; int priority; Node* next; Node(int x, int p) { data = x; priority = p; next = NULL; } }; // Return the value at head int peek(Node* head) { // Return the data of the node at the head of the list return head->data; } // Removes the element with the highest priority from the list Node* pop(Node* head) { // Store the current head node in a temporary variable Node* temp = head; // Move the head to the next node in the list head = head->next; // Free the memory of the removed head node delete temp; // Return the new head of the list return head; } // Function to push according to priority Node* push(Node* head, int d, int p) { Node* start = head; // Create new Node with the given data and priority Node* temp = new Node(d, p); // Special Case: Insert the new node before the head // if the list is empty or the head has lower priority if (head == NULL || head->priority > p) { // Insert the new node before the head temp->next = head; head = temp; } else { // Traverse the list to find the correct position // to insert the new node based on priority while (start->next != NULL && start->next->priority < p) { start = start->next; } // Insert the new node at the found position temp->next = start->next; start->next = temp; } return head; } // Function to check if the list is empty int isEmpty(Node* head) { return (head == NULL); } // Driver code int main() { Node* pq = new Node(4, 1); pq = push(pq, 5, 2); pq = push(pq, 6, 3); pq = push(pq, 7, 0); while (!isEmpty(pq)) { cout << " " << peek(pq); pq = pop(pq); } return 0; } // C code to implement Priority Queue // using Linked List #include <stdio.h> #include <stdlib.h> typedef struct node { int data; int priority; struct node* next; } Node; Node* newNode(int d, int p) { Node* temp = (Node*)malloc(sizeof(Node)); temp->data = d; temp->priority = p; temp->next = NULL; return temp; } // Return the value at head int peek(Node* head) { // Return the data of the node at the head of the list return head->data; } // Removes the element with the highest priority from the list Node* pop(Node* head) { // Store the current head node in a temporary variable Node* temp = head; // Move the head to the next node in the list head = head->next; // Free the memory of the removed head node free(temp); // Return the new head of the list return head; } // Function to push according to priority Node* push(Node* head, int d, int p) { Node* start = head; // Create new Node with the given data and priority Node* temp = newNode(d, p); // Special Case: Insert the new node before the head // if the list is empty or the head has lower priority if (head->priority > p) { // Insert the new node before the head temp->next = head; head = temp; } else { // Traverse the list to find the correct position // to insert the new node based on priority while (start->next != NULL && start->next->priority < p) { start = start->next; } // Insert the new node at the found position temp->next = start->next; start->next = temp; } return head; } // Function to check if the list is empty int isEmpty(Node* head) { return (head == NULL); } // Driver code int main() { Node* pq = newNode(4, 1); pq = push(pq, 5, 2); pq = push(pq, 6, 3); pq = push(pq, 7, 0); while (!isEmpty(pq)) { printf("%d ", peek(pq)); pq = pop(pq); } return 0; } // Java code to implement Priority Queue // using Linked List import java.util.* ; class GfG { static class Node { int data; int priority; Node next; } static Node node = new Node(); // Function to Create A New Node static Node newNode(int d, int p) { Node temp = new Node(); temp.data = d; temp.priority = p; temp.next = null; return temp; } // Return the value at head static int peek(Node head) { return (head).data; } // Removes the element with the // highest priority from the list static Node pop(Node head) { (head) = (head).next; return head; } // Function to push according to priority static Node push(Node head, int d, int p) { Node start = (head); // Create new Node Node temp = newNode(d, p); // Special Case: The head of list has lesser // priority than new node. So insert new // node before head node and change head node. if ((head).priority > p) { // Insert New Node before head temp.next = head; (head) = temp; } else { // Traverse the list and find a // position to insert new node while (start.next != null && start.next.priority <= p) { start = start.next; } // Either at the ends of the list // or at required position temp.next = start.next; start.next = temp; } return head; } // Function to check is list is empty static int isEmpty(Node head) { return ((head) == null)?1:0; } // Driver code public static void main(String args[]) { Node pq = newNode(4, 1); pq =push(pq, 5, 2); pq =push(pq, 6, 3); pq =push(pq, 7, 0); while (isEmpty(pq)==0) { System.out.printf("%d ", peek(pq)); pq=pop(pq); } } } # Python3 code to implement Priority Queue # using Singly Linked List class PriorityQueueNode: def __init__(self, value, pr): self.data = value self.priority = pr self.next = None # Global variable for the front of the queue front = None # Method to check Priority Queue is Empty def isEmpty(): return front is None # Method to add items in Priority Queue # According to their priority value def push(value, priority): global front # Condition check for checking Priority # Queue is empty or not if isEmpty(): front = PriorityQueueNode(value, priority) return 1 else: # Special condition check to see that # first node priority value if front.priority > priority: newNode = PriorityQueueNode(value, priority) newNode.next = front front = newNode return 1 else: temp = front while temp.next: if priority <= temp.next.priority: break temp = temp.next newNode = PriorityQueueNode(value, priority) newNode.next = temp.next temp.next = newNode return 1 # Method to remove high priority item # from the Priority Queue def pop(): global front if isEmpty(): return else: front = front.next return 1 # Method to return high priority node # value without removing it def peek(): if isEmpty(): return else: return front.data # Method to traverse the Priority Queue def traverse(): if isEmpty(): return "Queue is Empty!" else: temp = front while temp: print(temp.data, end=" ") temp = temp.next # Driver code if __name__ == "__main__": push(4, 1) push(5, 2) push(6, 3) push(7, 0) traverse() pop()
// C# code to implement Priority Queue // using Linked List using System; public class GfG { public class Node { public int data; public int priority; public Node next; } public static Node node = new Node(); static Node newNode(int d, int p) { Node temp = new Node(); temp.data = d; temp.priority = p; temp.next = null; return temp; } // Return the value at head static int peek(Node head) { return (head).data; } // Removes the element with the // highest priority from the list static Node pop(Node head) { (head) = (head).next; return head; } // Function to push according to priority static Node push(Node head, int d, int p) { Node start = (head); // Create new Node Node temp = newNode(d, p); // Special Case: The head of list // has lesser priority than new node. // So insert new node before head node // and change head node. if ((head).priority > p) { // Insert New Node before head temp.next = head; (head) = temp; } else { // Traverse the list and find a // position to insert new node while (start.next != null && start.next.priority <= p) { start = start.next; } // Either at the ends of the list // or at required position temp.next = start.next; start.next = temp; } return head; } // Function to check is list is empty static int isEmpty(Node head) { return ((head) == null) ? 1 : 0; } // Driver code public static void Main(string[] args) { Node pq = newNode(4, 1); pq = push(pq, 5, 2); pq = push(pq, 6, 3); pq = push(pq, 7, 0); while (isEmpty(pq) == 0) { Console.Write("{0:D} ", peek(pq)); pq = pop(pq); } } } // JavaScript code to implement Priority Queue // using Linked List class Node { constructor() { this.data = 0; this.priority = 0; this.next = null; } } var node = new Node(); function newNode(d, p) { var temp = new Node(); temp.data = d; temp.priority = p; temp.next = null; return temp; } // Return the value at head function peek(head) { // Return the data of the node at the head of the list return head.data; } // Removes the element with the highest priority from the list function pop(head) { // Store the current head node in a temporary variable var temp = head; // Move the head to the next node in the list head = head.next; // Return the new head of the list return head; } // Function to push according to priority function push(head, d, p) { var start = head; // Create new Node with the given data and priority var temp = newNode(d, p); // Special Case: Insert the new node before the head // if the list is empty or the head has lower priority if (head.priority > p) { // Insert New Node before head temp.next = head; head = temp; } else { // Traverse the list to find the correct position // to insert the new node based on priority while (start.next != null && start.next.priority <= p) { start = start.next; } // Insert the new node at the found position temp.next = start.next; start.next = temp; } return head; } // Function to check if the list is empty function isEmpty(head) { return head == null ? 1 : 0; } // Driver code var pq = newNode(4, 1); pq = push(pq, 5, 2); pq = push(pq, 6, 3); pq = push(pq, 7, 0); while (isEmpty(pq) == 0) { console.log(peek(pq) + " "); pq = pop(pq); } Time Complexity: push() -> O(n) , To insert an element we must traverse the list and find the proper position to insert the node . This makes the push() operation takes O(n) time.
pop -> O(1), as it is performed in constant time.
peek -> O(1), as it is performed in constant time.
Space Complexity: O(n), as we are making a List of size n