Cormen Edition 3 Exercise 10.2 Question 3 (Page No. 240)

Question

Implement a queue by a singly linked list $L$. The operations of $ENQUEUE$ and $DEQUEUE$ should still take $O(1)$ time.

Answer 1

Maintain two pointers: Head and Tail

DEQUEUE:

temp = new node();

tail->next = temp;

tail = temp;

ENQUEUE:

temp = head;

head = head->next;

delete(temp)

Answer 2

// A C program to demonstrate linked list based implementation of queue 
#include <stdio.h> 
#include <stdlib.h> 
  
// A linked list (LL) node to store a queue entry 
struct QNode { 
    int key; 
    struct QNode* next; 
}; 
  
// The queue, front stores the front node of LL and rear stores the 
// last node of LL 
struct Queue { 
    struct QNode *front, *rear; 
}; 
  
// A utility function to create a new linked list node. 
struct QNode* newNode(int k) 
{ 
    struct QNode* temp = (struct QNode*)malloc(sizeof(struct QNode)); 
    temp->key = k; 
    temp->next = NULL; 
    return temp; 
} 
  
// A utility function to create an empty queue 
struct Queue* createQueue() 
{ 
    struct Queue* q = (struct Queue*)malloc(sizeof(struct Queue)); 
    q->front = q->rear = NULL; 
    return q; 
} 
  
// The function to add a key k to q 
void enQueue(struct Queue* q, int k) 
{ 
    // Create a new LL node 
    struct QNode* temp = newNode(k); 
  
    // If queue is empty, then new node is front and rear both 
    if (q->rear == NULL) { 
        q->front = q->rear = temp; 
        return; 
    } 
  
    // Add the new node at the end of queue and change rear 
    q->rear->next = temp; 
    q->rear = temp; 
} 
  
// Function to remove a key from given queue q 
void deQueue(struct Queue* q) 
{ 
    // If queue is empty, return NULL. 
    if (q->front == NULL) 
        return; 
  
    // Store previous front and move front one node ahead 
    struct QNode* temp = q->front; 
  
    q->front = q->front->next; 
  
    // If front becomes NULL, then change rear also as NULL 
    if (q->front == NULL) 
        q->rear = NULL; 
  
    free(temp); 
} 
  
// Driver Program to test anove functions 
int main() 
{ 
    struct Queue* q = createQueue(); 
    enQueue(q, 10); 
    enQueue(q, 20); 
    deQueue(q); 
    deQueue(q); 
    enQueue(q, 30); 
    enQueue(q, 40); 
    enQueue(q, 50); 
    deQueue(q); 
    printf("Queue Front : %d \n", q->front->key); 
    printf("Queue Rear : %d", q->rear->key); 
    return 0; 
} 

Time Complexity: Time complexity of both operations enqueue() and dequeue() is O(1) as we only change few pointers in both operations. There is no loop in any of the operations.

Ref: https://www.geeksforgeeks.org/queue-linked-list-implementation/