思路：单调递减栈，找「上一个比当前价格大的元素索引」。正序遍历，弹出所有 ≤ 当前价格的索引；跨度 = 当前索引 - 栈顶索引（栈空则为 当前索引 + 1）。
There are 3 components to this project:
，更多细节参见新收录的资料
And it’s not just banking!，详情可参考新收录的资料
There's also Stream.broadcast() for push-based multi-consumer scenarios. Both require you to think about what happens when consumers run at different speeds, because that's a real concern that shouldn't be hidden.。新收录的资料是该领域的重要参考