---
tags: leetcode
---

# [622. Design Circular Queue](https://leetcode.com/problems/design-circular-queue/)
---
# My Solution

## The Key Idea for Solving This Coding Question

用一個陣列來實作環狀佇列 (circular queue) 。不過，==為了區分佇列為滿與佇列為空這兩種狀態，必須犧牲陣列中的一個元素不能存放資料==。所以我們的陣列的長度必須是 `k + 1` 。
我們必須用兩個索引 (index) 來操作這個陣列。這兩個索引分別命名為 `front` 與 `rear` 。 ==`front` 指向佇列的第一個元素==， ==`rear` 指向佇列最後一個元素的下一個元素==。所以， ==`rear` 所指向的那個元素永遠沒有資料==。這就是為什麼我們的陣列長度必須是 `k + 1` 的原因。
另外，實作 `Rear()` 時，對陣列的操作必須注意。當 `rear` 指向索引值 `0` 時，佇列尾端的元素位於陣列的索引值 `k` 的位置。不能只是單純的用 `rear - 1` 去操作陣列。否則，會得到一個負的索引值去操作陣列，這樣是會有 runtime error 出現的。

## C++ Code

class MyCircularQueue {
public:
    MyCircularQueue(int k) {
        front = 0;
        rear = 0;
        qLen = k + 1;
        q.resize(qLen);
    }
    
    bool enQueue(int value) {
        if (isFull()) {
            return false;
        }
        q[rear] = value;
        rear = (rear + 1) % qLen;
        
        return true;
    }
    
    bool deQueue() {
        if (isEmpty()) {
            return false;
        }
        front = (front + 1) % qLen;
        
        return true;
    }
    
    int Front() {
        if (isEmpty()) {
            return -1;
        }
        return q[front];
    }
    
    int Rear() {
        if (isEmpty()) {
            return -1;
        }
        return q[(rear - 1 + qLen) % qLen];
    }
    
    bool isEmpty() {
        return front == rear;
    }
    
    bool isFull() {
        return (rear + 1) % qLen == front;
    }
    
private:
    int front;
    int rear;
    int qLen;
    
    vector<int> q;
};

/**
 * Your MyCircularQueue object will be instantiated and called as such:
 * MyCircularQueue* obj = new MyCircularQueue(k);
 * bool param_1 = obj->enQueue(value);
 * bool param_2 = obj->deQueue();
 * int param_3 = obj->Front();
 * int param_4 = obj->Rear();
 * bool param_5 = obj->isEmpty();
 * bool param_6 = obj->isFull();
 */

## Time Complexity
* `MyCircularQueue`
$O(k)$
* `enQueue`
$O(1)$
* `deQueue`
$O(1)$
* `Front`
$O(1)$
* `Rear`
$O(1)$
* `isEmpty`
$O(1)$
* `isFull`
$O(1)$
## Space Complexity
* `MyCircularQueue`
$O(k)$
* `enQueue`
$O(1)$
* `deQueue`
$O(1)$
* `Front`
$O(1)$
* `Rear`
$O(1)$
* `isEmpty`
$O(1)$
* `isFull`
$O(1)$
# Miscellaneous
<!--
# Test Cases

["MyCircularQueue","enQueue","enQueue","enQueue","enQueue","Rear","isFull","deQueue","enQueue","Rear"]
[[3],[1],[2],[3],[4],[],[],[],[4],[]]

["MyCircularQueue","enQueue","Rear","Rear","deQueue","enQueue","Rear","deQueue","Front","deQueue","deQueue","deQueue"]
[[6],[6],[],[],[],[5],[],[],[],[],[],[]]

["MyCircularQueue","enQueue","enQueue","enQueue","enQueue","Rear","isFull","deQueue","enQueue","Rear","deQueue","deQueue","deQueue","isEmpty"]
[[3],[1],[2],[3],[4],[],[],[],[4],[],[],[],[],[]]

-->