LeetCode 622 Design Circular Queue

LeetCode

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:

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

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

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)