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This article will explain in detail how to solve the diving board problem with LeetCode. The editor thinks it is very practical, so I share it for you as a reference. I hope you can get something after reading this article.

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Topic description

Two sizes of boards are used to build diving boards, with sizes of longer and shorter respectively. A total of k blocks are used to calculate the lengths of all diving boards that can be obtained, and sort them from small to large. For example, shorter=1,longer=2,k=3, the final result is [3dy4, 5, 5, 6].

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Answer to the question

Idea: a total of k boards are used in mathematical calculation, so if longer uses I blocks, then shorter uses Kmuri blocks. Based on this basic idea, this problem can be completed in linear time, and the more shorter blocks are used, the shorter the overall diving board length is. Therefore, in the calculation, the number of shorter blocks decreases gradually from k, which can avoid the sorting operation later. In addition to the above basic cases, there are two special cases to consider, one is that when KFO, the return is empty, and the other is that when shorter=longer, the returned result is shorter*k (longer*k). Class Solution: def divingBoard (self, shorter: int, longer: int, k: int)-> List [int]: s = [] if krypton0: return s for i in range (0 Longer*i+shorter* 1): s.append (longer*i+shorter* (- iTunk)) s = list (set (s)) s.sort () return s on "how LeetCode solves the diving board problem" ends here. Hope that the above content can be helpful to you, so that you can learn more knowledge, if you think the article is good, please share it for more people to see.

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