232. Implement Queue using Stacks Easy

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

• void push(int x) Pushes element x to the back of the queue.
• int pop() Removes the element from the front of the queue and returns it.
• int peek() Returns the element at the front of the queue.
• boolean empty() Returns true if the queue is empty, false otherwise.

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, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations. 

Example 1:

Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false

Constraints:

• 1 <= x <= 9
• At most 100 calls will be made to push, pop, peek, and empty.
• All the calls to pop and peek are valid. 

Follow-up: Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.

解法1: push之前确保元素都在stack1（stack2是空的）， pop/peek之前确保元素都在stack2（stack1是空的）

时间复杂度 peek/pop/push 均为O(N)

class MyQueue {
    private Stack<Integer> stack1;
    private Stack<Integer> stack2;
    public MyQueue() {
        stack1 = new Stack();
        stack2 = new Stack();
    }
    public void push(int x) {
        while(!stack2.isEmpty()) stack1.push(stack2.pop());
        stack1.push(x);
    }
    public int pop() {
        while(!stack1.isEmpty()) stack2.push(stack1.pop());
        return stack2.pop();
    }
    public int peek() {
        while(!stack1.isEmpty()) stack2.push(stack1.pop());
        return stack2.peek();
    }
    public boolean empty() {
        return stack1.isEmpty() && stack2.isEmpty();
    }
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue obj = new MyQueue();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.peek();
 * boolean param_4 = obj.empty();
 */

优化后解法：push总是放到stack1，pop的时候先判定stack2是不是空的，如果是空的情况下才把stack1的元素都挪过来

时间复杂度 push为O(1), pop/peek amertized O(1)

class MyQueue {
    private Stack<Integer> stack1;
    private Stack<Integer> stack2;
    public MyQueue() {
        stack1 = new Stack();
        stack2 = new Stack();
    }
    public void push(int x) {
        stack1.push(x);
    }
    public int pop() {
        peek();
        return stack2.pop();
    }
    public int peek() {
        if(stack2.isEmpty()){
            while(!stack1.isEmpty()) stack2.push(stack1.pop());        
        }
        return stack2.peek();
    }
    public boolean empty() {
        return stack1.isEmpty() && stack2.isEmpty();
    }
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue obj = new MyQueue();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.peek();
 * boolean param_4 = obj.empty();
 */