LeetCode 232 - Implement Queue using Stacks

https://leetcode.com/problems/implement-queue-using-stacks

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).

https://discuss.leetcode.com/topic/17974/short-o-1-amortized-c-java-ruby

class MyQueue {

    Stack<Integer> input = new Stack();
    Stack<Integer> output = new Stack();
    
    public void push(int x) {
        input.push(x);
    }

    public void pop() {
        peek();
        output.pop();
    }

    public int peek() {
        if (output.empty())
            while (!input.empty())
                output.push(input.pop());
        return output.peek();
    }

    public boolean empty() {
        return input.empty() && output.empty();
    }
}

https://leetcode.com/articles/implement-queue-using-stacks/

X. Inefficent
https://discuss.leetcode.com/topic/28514/easy-java-solution-just-edit-push-method

I think your time complexity is worse than the solution with two stacks. In the two stack solution, each element is pushed and popped exactly twice, giving an amortized time complexity for push and popping O(1)

class MyQueue {
Stack<Integer> queue = new Stack<Integer>();
// Push element x to the back of queue.
public void push(int x) {
    Stack<Integer> temp = new Stack<Integer>();
    while(!queue.empty()){
        temp.push(queue.pop());
    }
    queue.push(x);
    while(!temp.empty()){
        queue.push(temp.pop());
    }
}

// Removes the element from in front of queue.
public void pop() {
    queue.pop();
}

// Get the front element.
public int peek() {
    return queue.peek();
}

// Return whether the queue is empty.
public boolean empty() {
    return queue.empty();
}

Method 2 (By making deQueue operation costly)

In this method, in en-queue operation, the new element is entered at the top of stack1. In de-queue operation, if stack2 is empty then all the elements are moved to stack2 and finally top of stack2 is returned.

enQueue(q,  x)
  1) Push x to stack1 (assuming size of stacks is unlimited).

deQueue(q)
  1) If both stacks are empty then error.
  2) If stack2 is empty
       While stack1 is not empty, push everything from satck1 to stack2.
  3) Pop the element from stack2 and return it. 

Method 2 is definitely better than method 1. Method 1 moves all the elements twice in enQueue operation, while method 2 (in deQueue operation) moves the elements once and moves elements only if stack2 empty.

void enQueue(struct queue *q, int x)

{

   push(&q->stack1, x);

}

  

/* Function to dequeue an item from queue */

int deQueue(struct queue *q)

{

   int x;

  

   /* If both stacks are empty then error */

   if(q->stack1 == NULL && q->stack2 == NULL)

   {

      printf("Q is empty");

      getchar();

      exit(0);

   }

   /* Move elements from satck1 to stack 2 only if

       stack2 is empty */

   if(q->stack2 == NULL)

   {

     while(q->stack1 != NULL)

     {

        x = pop(&q->stack1);

        push(&q->stack2, x);

     }

   }

   x = pop(&q->stack2);

   return x;

}

栈与队列的依赖实现

public class _06_MyQueue<T> {

    private Stack<T> stack1;

    private Stack<T> stack2;

    public _06_MyQueue() {

        stack1 = new Stack<T>();

        stack2 = new Stack<T>();

    }

    public void push(T item) {

        stack1.push(item);

    }

    public T pop() {

        if (stack2.size() > 0)

            return stack2.pop();

        while (stack1.size() > 0)

            stack2.push(stack1.pop());

        return stack2.pop();

    }

    public int size() {

        return stack1.size() + stack2.size();

    }

    public T peek() {

        if (stack2.size() > 0)

            return stack2.peek();

        while (stack1.size() > 0)

            stack2.push(stack1.pop());

        return stack2.peek();

    } 

}

Method 1 (By making enQueue operation costly)

This method makes sure that newly entered element is always at the top of stack 1, so that deQueue operation just pops from stack1. To put the element at top of stack1, stack2 is used.

enQueue(q, x)
  1) While stack1 is not empty, push everything from satck1 to stack2.
  2) Push x to stack1 (assuming size of stacks is unlimited).
  3) Push everything back to stack1.

dnQueue(q)
  1) If stack1 is empty then error
  2) Pop an item from stack1 and return it

Related: Leetcode - Implement Stack using Queues
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