Design Circular Queue LeetCode solution in C++

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class MyCircularQueue {
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  int head = -1;
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  int tail = -1;
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  int[] q;
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  int k;
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 public
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  MyCircularQueue(int k) {
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    q = new int[k];
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    this.k = k;
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  }
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 public
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  boolean enQueue(int value) {
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    if (head == -1 && tail == -1) {
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      // enqueue when the queue is empty
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      head = tail = 0;
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      q[tail] = value; // enQueue done
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      return true;
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    }
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    if ((tail == k - 1 && head == 0) || tail + 1 == head) {
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      // queue is full
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      return false;
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    }
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    tail++;
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    if (tail == k) // condition for circularity
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      tail = 0;
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    q[tail] = value; // enQueue done
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    return true;
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  }
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 public
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  boolean deQueue() {
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    if (head == -1) {
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      // check if q is already empty
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      return false;
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    }
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    if (head == tail) {
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      // only 1 element is there and head,tail both points same index
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      head = tail = -1; // deQueue done
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      return true;
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    }
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    head++;        // deQueue done
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    if (head == k) // condition for circularity
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      head = 0;
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    return true;
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  }
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 public
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  int Front() {
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    if (head == -1) return -1;
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    return q[head];
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  }
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 public
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  int Rear() {
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    if (tail == -1) return -1;
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    return q[tail];
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  }
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 public
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  boolean isEmpty() {
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    if (head == -1 && tail == -1) return true;
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    return false;
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  }
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 public
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  boolean isFull() {
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    if (tail == k - 1 || tail + 1 == head) return true;
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    return false;
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  }
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}
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/**
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 * Your MyCircularQueue object will be instantiated and called as such:
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 * MyCircularQueue obj = new MyCircularQueue(k);
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 * boolean param_1 = obj.enQueue(value);
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 * boolean param_2 = obj.deQueue();
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 * int param_3 = obj.Front();
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 * int param_4 = obj.Rear();
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 * boolean param_5 = obj.isEmpty();
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 * boolean param_6 = obj.isFull();
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 */