LeetCode 622 Design Circular Queue

LeetCode

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Design your implementation of the circular queue. The circular queue is a linear data structure in which the operations are performed based on FIFO (First In First Out) principle, and the last position is connected back to the first position to make a circle. It is also called "Ring Buffer".

One of the benefits of the circular queue is that we can make use of the spaces in front of the queue. In a normal queue, once the queue becomes full, we cannot insert the next element even if there is a space in front of the queue. But using the circular queue, we can use the space to store new values.

Implement the MyCircularQueue class:

• MyCircularQueue(k) Initializes the object with the size of the queue to be k.
• int Front() Gets the front item from the queue. If the queue is empty, return -1.
• int Rear() Gets the last item from the queue. If the queue is empty, return -1.
• boolean enQueue(int value) Inserts an element into the circular queue. Return true if the operation is successful.
• boolean deQueue() Deletes an element from the circular queue. Return true if the operation is successful.
• boolean isEmpty() Checks whether the circular queue is empty or not.
• boolean isFull() Checks whether the circular queue is full or not.

You must solve the problem without using the built-in queue data structure in your programming language.

Example 1:

<pre>Input ["MyCircularQueue", "enQueue", "enQueue", "enQueue", "enQueue", "Rear", "isFull", "deQueue", "enQueue", "Rear"] [[3], [1], [2], [3], [4], [], [], [], [4], []] Output [null, true, true, true, false, 3, true, true, true, 4]

Explanation MyCircularQueue myCircularQueue = new MyCircularQueue(3); myCircularQueue.enQueue(1); // return True myCircularQueue.enQueue(2); // return True myCircularQueue.enQueue(3); // return True myCircularQueue.enQueue(4); // return False myCircularQueue.Rear(); // return 3 myCircularQueue.isFull(); // return True myCircularQueue.deQueue(); // return True myCircularQueue.enQueue(4); // return True myCircularQueue.Rear(); // return 4 </pre>

Constraints:

• 1 <= k <= 1000
• 0 <= value <= 1000
• At most 3000 calls will be made to enQueue, deQueue, Front, Rear, isEmpty, and isFull.

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题目大意：

实现循环队列

解题思路Array(推荐)：

用两个指针为维持数据的开始和结束+1，由于循环队列，所以用mod来找到数组下标，也就是start和end一样时候，既可能是empty也可能是full，所以加一个变量记录

解题步骤：

用count就可以省去end和is_full两个变量
如果要实现同步队列，就引入self.queueLock = Lock(), with self.queueLock: 就进行enQueue操作

注意事项：

N/A

Python代码：

class MyCircularQueue:

    def __init__(self, k: int):
        self.data = [0] * k
        self.start = 0 # starting point of data
        self.end = 0 # ending point + 1 of data
        self.size = k
        self.is_full = False

    def enQueue(self, value: int) -> bool:
        if self.isFull():
            return False
        self.data[self.end] = value
        self.end = (self.end + 1) % self.size
        if self.start == self.end:
            self.is_full = True
        return True

    def deQueue(self) -> bool:
        if self.isEmpty():
            return False
        self.start = (self.start + 1) % self.size
        self.is_full = False
        return True

    def Front(self) -> int:
        if self.isEmpty():
            return -1
        return self.data[self.start]

    def Rear(self) -> int:
        if self.isEmpty():
            return -1
        return self.data[(self.end - 1) % self.size]

    def isEmpty(self) -> bool:
        return not self.is_full and self.start == self.end

    def isFull(self) -> bool:
        return self.is_full

算法分析：

时间复杂度为O(1)，空间复杂度O(n)

算法II解题思路Linked List：

比较简单，省略

算法分析：

时间复杂度为O(1)，空间复杂度O(n)