用队列实现栈操做

原题

　　Implement the following operations of a stack using queues.
　　push(x) – Push element x onto stack.
　　pop() – Removes the element on top of the stack.
　　top() – Get the top element.
　　empty() – Return whether the stack is empty.
　　Notes:
　　You must use only standard operations of a queue – which means only push to back, peek/pop from front, size, and is empty operations are valid.
　　Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
　　You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

题目大意

　　使用队列实现栈操做
　　push(x) – 元素入栈
　　pop() – 元素出栈
　　top() – 取栈顶元素值
　　empty() – 判断栈是否为空
　　注意：
　　只能使用队列的标准操做，先进先出，求队列元素数，判断队列是否为空
　　因为编程语言缘由，有些语言不支抚摩队列，可使用链表或双向链表代替，但仅能使用标准的队列操做
　　你能够假设全部的操做都是合法的，即：当队列为空时不会有元素出栈和求栈顶元素的操做

解题思路

　　用两个队列来模拟一个栈

代码实现

算法实现类

import java.util.LinkedList;
import java.util.List;

public class MyStack {

    // 维持两个队列，其中总有一个队列为空，为pop和top操做准备
    private List<Integer> aList = new LinkedList<>();
    private List<Integer> bList = new LinkedList<>();


    // Push element x onto stack.
    public void push(int x) {
        // 若是aList非空，就将x添加到aList中
        if (!aList.isEmpty()) {
            aList.add(x);
        }
        // 不然总添加到bList中
        else {
            bList.add(x);
        }
    }

    // Removes the element on top of the stack.
    public void pop() {

        // 两个队列中至少有一个为空，将aList设置非空
        if (aList.isEmpty()) {
            List<Integer> tmp = bList;
            bList = aList;
            aList = tmp;
        }

        // 除最后一个元素外都转移到bList中
        while (aList.size() > 1) {
            bList.add(aList.remove(0));
        }

        // 删除最后一个元素（对应就是入栈的栈顶元素）
        aList.clear();
    }

    // Get the top element.
    public int top() {

        // 两个队列中至少有一个为空，将aList设置非空
        if (aList.isEmpty()) {
            List<Integer> tmp = bList;
            bList = aList;
            aList = tmp;
        }

        // 除最后一个元素外都转移到bList中
        while (aList.size() > 1) {
            bList.add(aList.remove(0));
        }

        bList.add(aList.get(0));
        return aList.remove(0);
    }

    // Return whether the stack is empty.
    public boolean empty() {
        return aList.isEmpty() && bList.isEmpty();
    }
}