（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;
    }
    
}

（LeetCode）两个队列来实现一个栈

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