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Priority Queue Using Doubly Linked List in C++
We are given with the data and the priority as an integer value and the task is to create a doubly linked list as per the priority given and display the result.
Queue is a FIFO data structure in which the element which is inserted first is the first one to get removed. A Priority Queue is a type of queue in which elements can be inserted or deleted depending upon the priority. It can be implemented using queue, stack or linked list data structure. Priority queue is implemented by following these rules −
- Data or element with the highest priority will get executed before the data or element with the lowest priority.
- If two elements have the same priority than they will be executed in the sequence they are added in the list.
A node of a doubly linked list for implementing priority queue will contain three parts −
- Data − It will store the integer value.
- Next Address − It will store the address of a next node
- Previous Address − It will store the address of a previous node
- Priority − It will store the priority which is an integer value. It can range from 0-10 where 0 represents the highest priority and 10 represents the lowest priority.
Example
Input -
Output-
Algorithm
Start Step 1-> Declare a struct Node Declare info, priority Declare struct Node *prev, *next Step 2-> In function push(Node** fr, Node** rr, int n, int p) Set Node* news = (Node*)malloc(sizeof(Node)) Set news->info = n Set news->priority = p If *fr == NULL then, Set *fr = news Set *rr = news Set news->next = NULL Else If p <= (*fr)->priority then, Set news->next = *fr Set (*fr)->prev = news->next Set *fr = news Else If p > (*rr)->priority then, Set news->next = NULL Set (*rr)->next = news Set news->prev = (*rr)->next Set *rr = news Else Set Node* start = (*fr)->next Loop While start->priority > p Set start = start->next Set (start->prev)->next = news Set news->next = start->prev Set news->prev = (start->prev)->next Set start->prev = news->next Step 3-> In function int peek(Node *fr) Return fr->info Step 4-> In function bool isEmpty(Node *fr) Return (fr == NULL) Step 5-> In function int pop(Node** fr, Node** rr) Set Node* temp = *fr Set res = temp->info Set (*fr) = (*fr)->next free(temp) If *fr == NULL then, *rr = NULL Return res Step 6-> In function int main() Declare and assign Node *front = NULL, *rear = NULL Call function push(&front, &rear, 4, 3) Call function push(&front, &rear, 3, 2) Call function push(&front, &rear, 5, 2) Call function push(&front, &rear, 5, 7) Call function push(&front, &rear, 2, 6) Call function push(&front, &rear, 1, 4) Print the results obtained from calling the function pop(&front, &rear) Print the results obtained from calling the function peek(front) Stop
Example
#include <bits/stdc++.h> using namespace std; //doubly linked list node struct Node { int info; int priority; struct Node *prev, *next; }; //inserting a new Node void push(Node** fr, Node** rr, int n, int p) { Node* news = (Node*)malloc(sizeof(Node)); news->info = n; news->priority = p; // if the linked list is empty if (*fr == NULL) { *fr = news; *rr = news; news->next = NULL; } else { // If p is less than or equal front // node's priority, then insert the node // at front. if (p <= (*fr)->priority) { news->next = *fr; (*fr)->prev = news->next; *fr = news; } else if (p > (*rr)->priority) { news->next = NULL; (*rr)->next = news; news->prev = (*rr)->next; *rr = news; } else { // Finding the position where we need to // insert the new node. Node* start = (*fr)->next; while (start->priority > p) start = start->next; (start->prev)->next = news; news->next = start->prev; news->prev = (start->prev)->next; start->prev = news->next; } } } //the last value int peek(Node *fr) { return fr->info; } bool isEmpty(Node *fr) { return (fr == NULL); } int pop(Node** fr, Node** rr) { Node* temp = *fr; int res = temp->info; (*fr) = (*fr)->next; free(temp); if (*fr == NULL) *rr = NULL; return res; } // main function int main() { Node *front = NULL, *rear = NULL; push(&front, &rear, 4, 3); push(&front, &rear, 3, 2); push(&front, &rear, 5, 2); push(&front, &rear, 5, 7); push(&front, &rear, 2, 6); push(&front, &rear, 1, 4); printf("%d\n", pop(&front, &rear)); printf("%d\n", peek(front)); return 0; }
Output
5 3
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