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)。
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Раскрыты подробности похищения ребенка в Смоленске09:27
[&:first-child]:overflow-hidden [&:first-child]:max-h-full"
p->scavange= 0;