0225 - Implement Stack using Queues (Easy)

Problem Link

https://leetcode.com/problems/implement-stack-using-queues

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 element x 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 is empty, 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 to push, pop, top, and empty.
• All the calls to pop and top are valid.

Approach 1: 2 Queues

We can push all elements to one queue. For pop and top function, we move first $n - 1$ elements to another queue. What's left would be the top element. For pop function, we pop the top element as well and swap the queue.

Written by @wingkwong

class MyStack {
public:
    queue<int> q1, q2;
    MyStack() { }
    
    void push(int x) {
        q1.push(x);
    }
    
    int pop() {
        while (q1.size() > 1) {
            int x = q1.front();
            q1.pop();
            q2.push(x);
        }
        int res = q1.front();
        q1.pop();
        swap(q1, q2);
        return res;
    }
    
    int top() {
        while (q1.size() > 1) {
            int x = q1.front();
            q1.pop();
            q2.push(x);
        }
        return q1.front();
    }
    
    bool empty() {
        return q1.empty();
    }
};

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack* obj = new MyStack();
 * obj->push(x);
 * int param_2 = obj->pop();
 * int param_3 = obj->top();
 * bool param_4 = obj->empty();
 */

Approach 2: 1 Queue

For every push, we simply make the order backwards for push function. For pop() and top(), we can use front() to get the top element and return it.

Written by @wingkwong

class MyStack {
public:
    queue<int> q1;
    MyStack() { }
    
    void push(int x) {
        q1.push(x);
        for (int i = 1; i < q1.size(); i++) {
            q1.push(q1.front());
            q1.pop();
        }
    }
    
    int pop() {
        int x = q1.front(); q1.pop();
        return x;
    }
    
    int top() {
        return q1.front();
    }
    
    bool empty() {
        return q1.empty();
    }
};

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack* obj = new MyStack();
 * obj->push(x);
 * int param_2 = obj->pop();
 * int param_3 = obj->top();
 * bool param_4 = obj->empty();
 */