Leetcode: Implement Queue using Stacks

Question

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.
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, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
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Solution

code

class Queue(object):    def __init__(self):        """        initialize your data structure here.        """        self.stack1 = []        self.stack2 = []    def push(self, x):        """        :type x: int        :rtype: nothing        """        self.stack1.append(x)    def pop(self):        """        :rtype: nothing        """        if self.stack2!=[]:            self.stack2 = self.stack2[0:-1]        else:            while self.stack1!=[]:                temp = self.stack1[-1]                self.stack2.append(temp)                self.stack1 = self.stack1[0:-1]            if self.stack2!=[]:                self.stack2 = self.stack2[0:-1]    def peek(self):        """        :rtype: int        """        res = None        if self.stack2!=[]:            res = self.stack2[-1]            #self.stack2 = self.stack2[0:-1]        else:            while self.stack1!=[]:                temp = self.stack1[-1]                self.stack2.append(temp)                self.stack1 = self.stack1[0:-1]            if self.stack2!=[]:                res = self.stack2[-1]                #self.stack2 = self.stack2[0:-1]        return res    def empty(self):        """        :rtype: bool        """        return len(self.stack1)==0 and len(self.stack2)==0

Take home message

peek(): don not need to remove the front elements. All need to do is to return the value of the first element.