232. Implement Queue using Stacks

Problem description:

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.
Example:

MyQueue queue = new MyQueue();

queue.push(1);
queue.push(2);
queue.peek(); // returns 1
queue.pop(); // returns 1
queue.empty(); // returns false
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).

Solution:

This question is basically showing the understanding of data structure. We can use two stacks to help while we push into the stack. If you face this question in a interview, start with the difference with queue(FIFO) and stack(FILO).

class MyQueue {
public:
    /** Initialize your data structure here. */
    stack<int> stk;
    
    /** Push element x to the back of queue. */
    void push(int x) {
        stack<int> tmp;
        while(!stk.empty()){
            tmp.push(stk.top());
            stk.pop();
        }
        tmp.push(x);
        
        while(!tmp.empty()){
            stk.push(tmp.top());
            tmp.pop();
        }
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        int tmp= stk.top();
        stk.pop();
        return tmp;
    }
    
    /** Get the front element. */
    int peek() {
        return stk.top();
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        return stk.empty();
    }
};

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue obj = new MyQueue();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.peek();
 * bool param_4 = obj.empty();
 */

Time complexity:
push: $O(n)$, pop: $O(1)$, peek: $O(1)$, isEmpty: $O(1)$
Space Complexity: $O(n)$

reference:
https://goo.gl/4fv6iQ
https://goo.gl/KqsRJT