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0225 - Implement Stack using Queues (Easy)

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−1n - 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();
 */