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Implement Queue using Stacks

Problem Statement

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty`).

Implement the MyQueue class:

  • void push(int x) Pushes element x to the back of the queue.
  • int pop() Removes the element from the front of the queue and returns it.
  • int peek() Returns the element at the front of the queue.
  • boolean empty() Returns true if the queue is empty, `false otherwise.

Notes:

  • You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
  • Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.

Leetcode Link

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
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

Constraints:

  • 1 <= x <= 9
  • At most 100 calls will be made to push, pop, peek, and empty.
  • All the calls to pop and peek are valid.

Code

Python Code

class Queue(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.inStack, self.outStack = [], []

def push(self, x):
"""
:type x: int
:rtype: nothing
"""
self.inStack.append(x)

def pop(self):
"""
:rtype: nothing
"""
self.move()
self.outStack.pop()

def peek(self):
"""
:rtype: int
"""
self.move()
return self.outStack[-1]

def empty(self):
"""
:rtype: bool
"""
return (not self.inStack) and (not self.outStack)

def move(self):
"""
:rtype nothing
"""
if not self.outStack:
while self.inStack:
self.outStack.append(self.inStack.pop())