17:19, 27 февраля 2026Силовые структуры
for (int i = 1; i < 10; i++) {
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As you might expect, the result of this is that colours which lie closer to the input pixel are given a greater proportion of the total influence with ever-increasing values of . This is not mentioned in the cited paper but it might be nice to consider for your own implementation.
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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