binary-search

来源： https://stackoverflow.com/questions/39221303/binary-search-algorithm-implementations

来源： https://stackoverflow.com/questions/39221303/binary-search-algorithm-implementations

来源： https://stackoverflow.com/questions/39221303/binary-search-algorithm-implementations

来源： https://stackoverflow.com/questions/39221303/binary-search-algorithm-implementations

来源： https://stackoverflow.com/questions/63177247/python-fastest-way-to-subtract-elements-of-datasets-of-hdf5-file

来源： https://stackoverflow.com/questions/61420575/is-a-binary-search-really-better-than-my-solution-google-kickstart-2020-round

来源： https://stackoverflow.com/questions/61420575/is-a-binary-search-really-better-than-my-solution-google-kickstart-2020-round

问题 This question already has answers here : Time complexity of binary search for an unsorted array (3 answers) Closed 4 years ago . I have an array that looks like 2 6 8 5 34 1 12 Can I use a binary search on some subarray? 回答1: You can use binary search on only one kind of "unsorted" array - the rotated array. It can be done in O(log n) time like a typical binary search, but uses an adjusted divide and conquer approach. You can find a discussion about it here. 回答2: No, binary search needs a

问题 This question already has answers here : Time complexity of binary search for an unsorted array (3 answers) Closed 4 years ago . I have an array that looks like 2 6 8 5 34 1 12 Can I use a binary search on some subarray? 回答1: You can use binary search on only one kind of "unsorted" array - the rotated array. It can be done in O(log n) time like a typical binary search, but uses an adjusted divide and conquer approach. You can find a discussion about it here. 回答2: No, binary search needs a

问题 I have an array of true/false values with a very simple structure: # the real array has hundreds of thousands of items positions = np.array([True, False, False, False, True, True, True, True, False, False, False], dtype=np.bool) I want to traverse this array and output the places where changes happen (true becomes false or the contrary). For this purpose, I've put together two different approaches: a recursive binary search (see if all values are the same, if not, split in two, then recurse)