LeetCode--Implement Queue using Stacks-蒲公英云

LeetCode--Implement Queue using Stacks

Problem:

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.
Notes:
- You must use onlystandard 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).

Analysis:
题意：用栈实现队列
用两个堆栈实现，一个放push的数据，另一个放pop的数据，但是在pop在前需要将push栈中的数据push到pop栈中,这样就可以使pop先进先出了。
注意：

堆栈的输入元素是对象，所以需要定义Integer类型
pop时应该先pop堆栈（pop的），如果为空就要将push堆栈的数据push到pop堆栈中，然后再pop出去。
peek需要返回int类型，所以注意输出时return，不要忘了。
判断是否为空主要是判断两个栈是否同时为空。

Anwser:

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