Pokémon Presents 2026: All the news and trailers

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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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for (int i = n - 1; i 0; i--) {。关于这个话题,WPS下载最新地址提供了深入分析

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