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

// Author : Tianming Cao

// Date : 2018-02-02

/**********************************************************************************

* 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).

*

**********************************************************************************/

package myQueue;

import java.util.Stack;

/**

* This problem is a sibling of problem 225(https://leetcode.com/problems/implement-stack-using-queues/description/)

* The solution is:

* 1. stack1 is always in charge of push operation

* 2. stack2 is always in charge of peek and pop operation

* 3. if we want to do peek or pop operation, but stack2 is empty,

* we should first pop all the elements from stack1 and push them into stack2 in turn.

* Give a Example:

*

* First, push numbers "1,2,3,4,5" to stack1, then stack1's structure is:

*

* |5|

* |4|

* |3|

* |2|

* |1|

*

* Second, if we want to get the front element "1",we should pop all the elements of stack1 and push them into stack2,

* after this, stack1 is empty, and stack2's structrue is:

*

* |1|

* |2|

* |3|

* |4|

* |5|

*

* So we can get stack2's top element "1" as the front element of queue.

*

* Next, if we want to push "6" to the back of queue, we should push "6" into stack1 as before, so stack1's structure is:

*

* |6|

*

* Finally, if we want to do pop operation twice ,we should remove the top element of stack2 twice, so stack2's structure is:

*

* |3|

* |4|

* |5|

*

* as expect, the removed element is "1" and "2".

*/

public class MyQueue {

public Stack<Integer> stack1;

public Stack<Integer> stack2;

public int size;

public MyQueue() {

stack1 = new Stack<>();

stack2 = new Stack<>();

size = 0;

}

public void push(int x) {

stack1.push(x);

size++;

}

public int pop() {

if (stack2.isEmpty()) {

while (!stack1.isEmpty()) {

stack2.push(stack1.pop());

}

}

int value = stack2.pop();

size--;

return value;

}

public int peek() {

if (stack2.isEmpty()) {

while (!stack1.isEmpty()) {

stack2.push(stack1.pop());

}

}

return stack2.peek();

}

public boolean empty() {

return size == 0 ? true : false;

}

}