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.

- 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).

**Understand the problem:**

The problem can be solved by using two stacks, stack1 and stack2.

-- push(x), push an element into stack1. O(1) time complexity.

-- pop(), check if the stack 2 is empty. If not, pop from stack2. If yes, pop all elements from stack1 and push into stack2. Then pop from stack2. Amortized time complexity is O(1).

-- peek(). The same as pop().

-- isEmpty(). Check if both stack1 and stack2 are empty.

**Code (Java):**

class MyQueue { private Stack<Integer> stack1; private Stack<Integer> stack2; public MyQueue() { this.stack1 = new Stack<Integer>(); this.stack2 = new Stack<Integer>(); } // Push element x to the back of queue. public void push(int x) { stack1.push(x); } // Removes the element from in front of queue. public void pop() { if (!stack2.isEmpty()) { stack2.pop(); } else { while (!stack1.isEmpty()) { stack2.push(stack1.pop()); } stack2.pop(); } } // Get the front element. public int peek() { int ret = 0; if (!stack2.isEmpty()) { ret = stack2.peek(); } else { while (!stack1.isEmpty()) { stack2.push(stack1.pop()); } ret = stack2.peek(); } return ret; } // Return whether the queue is empty. public boolean empty() { return stack1.isEmpty() && stack2.isEmpty(); } }

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