Implement Queue by Two 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.

You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

Analysis

Make use of the nature that two opeartions of in/out stack opeartions can get the order require by a FIFO order.

For example, push(1), push(2), push(3), result in {1, 2, 3} in stack, where 3 is the stack top.

However, push it to another stack result in {3, 2, 1}, which is the desired FIFO order.

Complexity

Amortized cost: O(1), since each element is only moved once from inStack to outStack.

Code

class MyQueue {
  private Stack<Integer> inStack = new Stack<>();
  private Stack<Integer> outStack = new Stack<>();

  // Push element x to the back of queue.
  public void push(int x) {
    inStack.push(x);
  }

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

  // Get the front element.
  public int peek() {
    if (outStack.isEmpty()) {
      moveFromInStackToOutStack();
    }
    return outStack.peek();
  }

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

  // pop all elements from inStack into outStack.
  private void moveFromInStackToOutStack() {
    while (!inStack.isEmpty()) {
      outStack.push(inStack.pop());
    }
  }
}