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)。
actual fun fromByteArray(byteArray: ByteArray): PlatformByteArray {
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I am generally cynical about anything foisted upon us by the game’s overlords, but after a brilliant couple of nights of football Uefa must be delighted with the drama and excitement these playoffs produced.。同城约会是该领域的重要参考
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Retroactive Privilege Expansion. You created a Maps key three years ago and embedded it in your website's source code, exactly as Google instructed. Last month, a developer on your team enabled the Gemini API for an internal prototype. Your public Maps key is now a Gemini credential. Anyone who scrapes it can access your uploaded files, cached content, and rack up your AI bill. Nobody told you.