leetcode-232-easy

Implement Queue using Stacks

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.

思路一：想到的是笨办法，pop 和 peek 操作直接倒两个栈。看了一下官方实现，有更简单的操作，直接用另一个栈 pop 和 peek，由于队列先入先出的特点，所以不会影响到 push 操作

    private Stack<Integer> stackOne;
    private Stack<Integer> stackTwo;

    public MyQueue() {
        stackOne = new Stack<>();
        stackTwo = new Stack<>();
    }

    public void push(int x) {
        stackOne.push(x);
    }

    public int pop() {
        while (!stackOne.isEmpty()) {
            stackTwo.push(stackOne.pop());
        }

        int val = stackTwo.pop();

        while (!stackTwo.isEmpty()) {
            stackOne.push(stackTwo.pop());
        }

        return val;
    }

    public int peek() {
        while (!stackOne.isEmpty()) {
            stackTwo.push(stackOne.pop());
        }

        int val = stackTwo.peek();

        while (!stackTwo.isEmpty()) {
            stackOne.push(stackTwo.pop());
        }

        return val;
    }

    public boolean empty() {
        return stackOne.isEmpty();
    }