描述：nums1 中数字 x 的「下一个更大元素」是指 x 在 nums2 中对应位置右侧的第一个比 x 大的元素。给你两个没有重复元素的数组 nums1 和 nums2，其中 nums1 是 nums2 的子集。对于每个 nums1[i]，找出其在 nums2 中的下一个更大元素；若不存在，返回 -1。
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。关于这个话题，搜狗输入法2026提供了深入分析
Цены на нефть взлетели до максимума за полгода17:55
The semantics around releasing locks with pending reads were also unclear for years. If you called read() but didn't await it, then called releaseLock(), what happened? The spec was recently clarified to cancel pending reads on lock release – but implementations varied, and code that relied on the previous unspecified behavior can break.