Queue¶
Table of Contents¶
232. Implement Queue using Stacks¶
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LeetCode | LeetCode CH (Easy)
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Tags: stack, design, queue
- Implement the following operations of a queue using stacks.
push(x)- Push element x to the back of queue.pop()- Removes the element from in front of queue.peek()- Get the front element.empty()- Return whether the queue is empty.
232. Implement Queue using Stacks - Python Solution
class MyQueue:
def __init__(self):
self.stack_in = []
self.stack_out = []
def push(self, x: int) -> None:
self.stack_in.append(x)
def pop(self) -> int:
if self.empty():
return None
if self.stack_out:
return self.stack_out.pop()
else:
for _ in range(len(self.stack_in)):
self.stack_out.append(self.stack_in.pop())
return self.stack_out.pop()
def peek(self) -> int:
answer = self.pop()
self.stack_out.append(answer)
return answer
def empty(self) -> bool:
return not (self.stack_in or self.stack_out)
obj = MyQueue()
obj.push(1)
print(obj.pop()) # 1
print(obj.peek()) # None
print(obj.empty()) # False
225. Implement Stack using Queues¶
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LeetCode | LeetCode CH (Easy)
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Tags: stack, design, queue
225. Implement Stack using Queues - Python Solution
from collections import deque
# Queue
class MyStack:
def __init__(self):
self.q1 = deque() # main queue
self.q2 = deque() # auxiliary queue
def push(self, x: int) -> None:
self.q1.append(x)
def pop(self) -> int:
while len(self.q1) > 1:
self.q2.append(self.q1.popleft())
res = self.q1.popleft()
self.q1, self.q2 = self.q2, self.q1
return res
def top(self) -> int:
while len(self.q1) > 1:
self.q2.append(self.q1.popleft())
res = self.q1[0]
self.q2.append(self.q1.popleft())
self.q1, self.q2 = self.q2, self.q1
return res
def empty(self) -> bool:
return not self.q1
obj = MyStack()
obj.push(1)
obj.push(2)
obj.push(3)
obj.push(4)
print(obj.pop()) # 4
print(obj.top()) # 3
print(obj.empty()) # False
print(obj.pop()) # 3