Implement Stacks using Queue
Problem Statement
Implement a last-in-first-out (LIFO
) stack using only two queues. The implemented stack should support all the functions of a normal stack (push
, top
, pop
, and empty
).
Implement the MyStack
class:
void push(int x)
Pushes elementx
to the top of the stack.int pop()
Removes the element on the top of the stack and returns it.int top()
Returns the element on the top of the stack.boolean empty()
Returns true if the stack isempty
,false
otherwise.
Notes:
- You must use only standard operations of a queue, which means that only push to back, peek/pop from front, size and is empty operations are valid.
- Depending on your language, the queue may not be supported natively. You may simulate a queue using a list or deque (double-ended queue) as long as you use only a queue's standard operations.
Example 1:
Input
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 2, 2, false]
Explanation
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.empty(); // return False
Constraints:
1 <= x <= 9
- At most
100
calls will be made topush
,pop
,top
, andempty
. - All the calls to
pop
andtop
are valid.
Code
Python Code
from Queue import Queue
class MyStack(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.q1, self.q2 = Queue(), Queue()
return
def push(self, x):
"""
:type x: int
:rtype: nothing
"""
self.q1.put(x)
return
def pop(self):
"""
:rtype: nothing
"""
while self.q1.qsize() > 1:
x = self.q1.get()
self.q2.put(x)
if self.q1.qsize() == 1:
result = self.q1.get()
self.q1, self.q2 = self.q2, self.q1
return result
def top(self):
"""
:rtype: int
"""
while self.q1.qsize() > 1:
x = self.q1.get()
self.q2.put(x)
if self.q1.qsize() == 1:
result = self.q1.get()
self.q2.put(result)
self.q1, self.q2 = self.q2, self.q1
return result
def empty(self):
"""
:rtype: bool
"""
return not bool(self.q1.qsize() + self.q2.qsize())