2 queue as 1 stack, 2 stack as 1 queue
另外这个hexo有个bug，经常会说landscape主题找不到，其实我都没有用这个主题，需要推倒hexo根目录上hexo clean一下 再hexo g, hexo d, 估计是tommy513写的有个bug，最后自己也没修复掉，
要是有人能找到问题原因还挺不错的~~~
2个栈边队列，设为A B
入队列： 入A
出队列： 如果|A|>0,pop A 到push B直到A只有一个元素，然后输出Ａ达到先进先出，那么如果A空了，就直接pop B， 如果B也空了
if A 不空： pop A 到push B直到A只有一个元素，然后popA中的这个元素
else if B: 不空，popB
else:
栈已空
感觉出入队列至少有一个O(n)的，不知道有没有O(1)，都已经加了空间代价按道理有的。
2个队列变栈，设为A B
入栈：入A
出栈：(还没理清楚，感觉好像挺麻烦的，这个有通用的思路么)
由于队列倒来倒去顺序还是不变，所以比栈操作会更多一些，because pop and push stack can reverse the sequences.
所以队列是这样的，如果A不空，把A n-1个点逐个dequeue然后enqueue到B, 然后出第N个作为pop，如果A空，那么把B n-1 个点dequeue然后enqueue到A，然后第N个作为pop，一定是先看A，如果A空了再看B，因为入”栈”是从A入的
if n=|A|>=1，dequeue A n-1 to B, dequeue A nth element 作为 "pop"
else if |B|>=1 dequeue B n-1 to A, dequeue B nth element 作为 "pop"
else
can not pop as stack empty
这种DS设计逻辑感觉没啥可以总结出的东西，没啥规律，有点靠灵光一现的感觉。。。感觉就是写几个case，然后归纳出一套处理的流程的规则之类的。。。。
现在想清楚了，总结一下，入的话都简单，选定一个例如A。出的时候根据DS 因地制宜，例如2 stack to be queue, 首先一开始如5个，然后”dequeue”, 将A n-1 pop push to B, 然后 pop A as “dequeue”, if “enqueue” 6, 6 should as the last,
so we can not first choose A for dequeue, instead choose B, if B empty, choose A. And must first if A, then else if B, can not effect the sequence to first if B, then else if A as data first into A then to B.so algorithm would be:
stack A B as queue
enqueue: push A
dequeue:
if A not empty, pop A n-1 element to B, pop A nth element as "dequeue" // first would be all into A, as into "stack" or "queue", so element would first be A, then to B
else if B not empty, pop B as "dequeue" //as when A element to B, its sequence reversed and fit for dequeue "first in first out", so directly pop B would be "dequeue"
else
"queue" empty, can not dequeue
queue A B as stack
push: enqueue A
pop: if n=|A|>=1，dequeue A n-1 to B, dequeue A nth element as "pop"// as first into A
else if |B|>=1 dequeue B n-1 to A, dequeue B nth element as "pop"// queue is first in first out, and dequeue then enqueue would not effect its sequence
else
can not pop as "stack" empty