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LeetCode_232.py
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"""
请你仅使用两个栈实现先入先出队列。队列应当支持一般队列支持的所有操作(push、pop、peek、empty):
实现 MyQueue 类:
void push(int x) 将元素 x 推到队列的末尾
int pop() 从队列的开头移除并返回元素
int peek() 返回队列开头的元素
boolean empty() 如果队列为空,返回 true ;否则,返回 false
说明:
你 只能 使用标准的栈操作 —— 也就是只有push to top,peek/pop from top,size, 和is empty操作是合法的。
你所使用的语言也许不支持栈。你可以使用 list 或者 deque(双端队列)来模拟一个栈,只要是标准的栈操作即可。
输入:
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
输出:
[null, null, null, 1, 1, false]
解释:
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false
"""
class MyQueue:
def __init__(self):
self.stack1 = []
self.stack2 = []
"""
Initialize your data structure here.
"""
def push(self, x: int) -> None:
"""
Push element x to the back of queue.
"""
self.stack1.append(x)
def pop(self) -> int:
"""
Removes the element from in front of queue and returns that element.
"""
if len(self.stack2) == 0 and len(self.stack1) == 0:
return None
elif len(self.stack2) == 0 and len(self.stack1) >= 0:
for i in range(0, len(self.stack1)):
self.stack2.append(self.stack1.pop(-1))
return self.stack2.pop(-1)
else:
return self.stack2.pop(-1)
def peek(self) -> int:
if len(self.stack2) == 0 and len(self.stack1) == 0:
return None
elif len(self.stack2) == 0 and len(self.stack1) >= 0:
for i in range(0, len(self.stack1)):
self.stack2.append(self.stack1.pop(-1))
tmp = self.stack2.pop(-1)
self.stack2.append(tmp)
return tmp
else:
tmp = self.stack2.pop(-1)
self.stack2.append(tmp)
return tmp
"""
Get the front element.
"""
def empty(self) -> bool:
"""
Returns whether the queue is empty.
"""
if len(self.stack2) == 0 and len(self.stack1) == 0:
return True
else:
return False
if __name__ == '__main__':
myQueue = MyQueue()
myQueue.push(1)
myQueue.push(2)
print(myQueue.peek())
print(myQueue.pop())
print(myQueue.empty())