题目：原题链接（中等）
标签：队列
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | 所有操作 = O ( 1 ) O(1) O(1) | O ( K ) O(K) O(K) | 80ms (89.72%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。
解法一（使用定长列表实现）：
class MyCircularQueue:
def __init__(self, k: int):
"""
Initialize your data structure here. Set the size of the queue to be k.
"""
self.queue = [0] * k
self.index = 0
self.count = 0
self.size = k
def enQueue(self, value: int) -> bool:
"""
Insert an element into the circular queue. Return true if the operation is successful.
"""
if self.count < self.size:
self.queue[(self.index + self.count) % self.size] = value
self.count += 1
return True
else:
return False
def deQueue(self) -> bool:
"""
Delete an element from the circular queue. Return true if the operation is successful.
"""
if self.count > 0:
self.index = (self.index + 1) % self.size
self.count -= 1
return True
else:
return False
def Front(self) -> int:
"""
Get the front item from the queue.
"""
if self.count > 0:
return self.queue[self.index]
else:
return -1
def Rear(self) -> int:
"""
Get the last item from the queue.
"""
if self.count > 0:
return self.queue[(self.index + self.count - 1) % self.size]
else:
return -1
def isEmpty(self) -> bool:
"""
Checks whether the circular queue is empty or not.
"""
return self.count == 0
def isFull(self) -> bool:
"""
Checks whether the circular queue is full or not.
"""
return self.count == self.size
