Implement a queue using two stacks

The space cost is O(n). The runtime cost of enqueue is O(1); the runtime cost of
dequeue is O(n). Using the "accounting method", the cost of an item in the
system is O(1). Here's why:

+------------+----------------------------+------+
| enqueue    | push onto lhs              | O(1) |
+------------+----------------------------+------+
| lhs -> rhs | pop off lhs; push onto rhs | O(1) |
+------------+----------------------------+------+
| dequeue    | pop off rhs                | O(1) |
+------------+----------------------------+------+

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