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Published byJosiah Kirkby Modified over 3 years ago

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1 Array-based Implementation An array Q of maximum size N Need to keep track the front and rear of the queue: f: index of the front object r: index immediately past the rear element Note: Q[r] is empty (does not store any object)

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2 Array-based Implementation Front element: Q[f] Rear element: Q[r – 1] Queue is empty: f = r Queue size: r – f

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3 Dequeue() and Enqueue() Algorithm dequeue(): if (isEmpty()) throw QueueEmptyException; temp = Q[f]; f = f + 1; return temp; Algorithm enqueue(object): if (r == N) throw QueueFullException; Q[r] = object; r = r + 1;

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4 Circular Array Implementation Analogy: A snake chases its tail Front element: Q[f] Rear element: Q[r – 1] Incrementing f, r f = (f + 1) mod N r = (r + 1) mod N mod: Java operator “%”

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5 Circular Array Implementation Queue size = (N – f + r) mod N → verify this Queue is empty: f = r When r reaches and overlaps with f, the queue is full: r = f To distinguish between empty and full states, we impose a constraint: Q can hold at most N – 1 objects (one cell is wasted). So r never overlaps with f, except when the queue is empty.

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6 Pseudo-code Algorithm enqueue(object): if (size() == N – 1) throw QueueFullException; Q[r] = object; r = (r + 1) mod N; Algorithm dequeue(): if (isEmpty()) throw QueueEmptyException; temp = Q[f]; f = (f + 1) mod N; return temp;

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7 Pseudo-code Algorithm front(): if (isEmpty()) throw QueueEmptyException; return Q[f]; Algorithm isEmpty(): return (f = r); Algorithm size(): return ((N – f + r) mod N); Homework: Remove the constraint “Q can hold at most N – 1 objects”. That is, Q can store up to N objects. Implement the Queue ADT using a circular array. Note: there is no corresponding built-in Java class for queue ADT

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8 Double-Ended Queue ADT Deque (pronounced “deck”) Allows insertion and deletion at both the front and the rear of the queue Deque ADT: operations addFirst(e): insert e at the beginning of the deque addLast(e): insert e at the end of the deque removeFirst(): remove and return the first element removeLast(): remove and return the last element getFirst():return the first element getLast():return the last element isEmpty(): return true if deque is empty; false otherwise size():return the number of objects in the deque

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9 Implementation Choices Array (homework) Singly linked list: removing at the tail costs θ(n) Doubly linked list

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10 removeLast() and addLast()

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11 Implementing Stacks and Queues with Deques

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