题意：双栈实现队列；要求每个入队、出队操作均摊O(1)复杂度
题解：
用一个栈in维护入队元素，另一个栈out维护出队元素
出队或取队头元素：首先判断栈out是否为空，如果为空，将栈in中的元素pop()到栈out中，那么栈out栈顶元素即为原队列队头元素。（米奇妙妙屋啊～）
判断队空：栈in和栈out都空时队空
Java代码
class MyQueue {
Stack<Integer> in = new Stack<>();
Stack<Integer> out = new Stack<>();
public MyQueue() {
}
public void push(int x) {
in.push(x);
}
public int pop() {
if(out.empty()) {
inToOut();
}
return out.pop();
}
public int peek() {
if(out.empty()) {
inToOut();
}
return out.peek();
}
public boolean empty() {
return in.empty() && out.empty();
}
public void inToOut() {
while(!in.empty()) {
out.push(in.pop());
}
}
}
/**
* Your MyQueue object will be instantiated and called as such:
* MyQueue obj = new MyQueue();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.peek();
* boolean param_4 = obj.empty();
*/