LeetCode Implement Queue using Stacks

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).

思路分析：这题和几年前写过的一篇面经博文是相同的题目： 面试题研究 用两个栈模拟实现队列。基本做法很简单，用两个栈就可以模拟一个队列，基本思路是两次后进先出 = 先进先出，元素入队列总是入A栈，元素出队列如果B栈不为空直接弹出B栈头元素；如果B栈为空就把A栈元素出栈全部压入B栈，再弹出B栈头，这样就模拟出了一个队列。核心就是保证每个元素出栈时都经过了A，B两个栈，这样就实现了两次后进先出=先进先出。

AC Code

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