（LeetCode）两个队列来实现一个栈

原题如下：

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).

基本思路是：假设有两个队列Q1和Q2，当二者都为空时，入栈操作可以用入队操作来模拟，可以随便选一个空队列，假设选Q1进行入栈操作，现在假设a,b,c依次入栈了（即依次进入队列Q1），这时如果想模拟出栈操作，则需要将c出栈，因为在栈顶，这时候可以考虑用空队列Q2，将a，b依次从Q1中出队，而后进入队列Q2，将Q1的最后一个元素c出队即可，此时Q1变为了空队列，Q2中有两个元素，队头元素为a，队尾元素为b，接下来如果再执行入栈操作，则需要将元素进入到Q1和Q2中的非空队列，即进入Q2队列，出栈的话，就跟前面的一样，将Q2除最后一个元素外全部出队，并依次进入队列Q1，再将Q2的最后一个元素出队即可。

Java实现代码如下：

class MyStack {LinkedList<Integer> queue1 = new LinkedList<Integer>();LinkedList<Integer> queue2 = new LinkedList<Integer>();    // Push element x onto stack.    public void push(int x) {        if(queue1.size()==0&&queue2.size()==0)        {        queue1.offer(x);        }        else if(queue1.size()==0)        {        queue2.offer(x);        }        else        {        queue1.offer(x);        }    }    // Removes the element on top of the stack.    public void pop() {    if(queue1.size()!= 0)    {    int length = queue1.size();        for(int i =0;i<length-1;i++)        {        queue2.offer(queue1.poll());        }        queue1.poll();    }    else    {    int length = queue2.size();    for(int i =0;i<length-1;i++)    {    queue1.offer(queue2.poll());    }     queue2.poll();    }    }    // Get the top element.    public int top() {    if(queue1.size()!= 0)    {    int length = queue1.size();        for(int i =0;i<length-1;i++)        {        queue2.offer(queue1.poll());        }        int result =  queue1.element();        queue2.offer(queue1.poll());return result;    }    else    {    int length = queue2.size();    for(int i =0;i<length-1;i++)    {    queue1.offer(queue2.poll());    }    int result = queue2.element();    queue1.offer(queue2.poll());    return result;    }    }    // Return whether the stack is empty.    public boolean empty() {        return queue1.size()==0&&queue2.size()==0;    }    }