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)。
There are many topics we haven't covered: interrupts, exceptions, task switching, and seldom-visited corners like call gates. I'll try to address them in future posts.
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Wordle today: Answer, hints for February 27, 2026
But of course, hard coding a size guess is a bit rigid.