思路：先对 nums2 用单调栈求每个元素的下一个更大值，存入 Map 缓存；再遍历 nums1 直接查 Map 得结果。时间复杂度 O(len1 + len2)。
How to watch the NFL online for free
。关于这个话题，同城约会提供了深入分析
text += dec.decode();
Not all streaming workloads involve I/O. When your source is in-memory and your transforms are pure functions, async machinery adds overhead without benefit — you're paying for coordination of "waiting" that adds no benefit.