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* Retranslate the complete English edition from Simplified Chinese * Polish three English explanations after reader-focused review
64 lines
1.9 KiB
Ruby
64 lines
1.9 KiB
Ruby
=begin
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File: binary_search.rb
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Created Time: 2024-04-09
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Author: Blue Bean (lonnnnnnner@gmail.com)
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=end
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### Binary search (closed interval) ###
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def binary_search(nums, target)
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# Initialize the closed interval [0, n-1], with i and j pointing to the first and last elements
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i, j = 0, nums.length - 1
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# Loop until the search interval is empty (when i > j)
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while i <= j
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# Ruby integers can grow arbitrarily large, limited by available memory, so integer overflow is not a concern here
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m = (i + j) / 2 # Calculate the midpoint index m
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if nums[m] < target
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i = m + 1 # In this case, target must be in [m+1, j] if it exists
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elsif nums[m] > target
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j = m - 1 # In this case, target must be in [i, m-1] if it exists
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else
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return m # The target element was found; return its index
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end
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end
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-1 # The target element was not found; return -1
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end
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### Binary search (left-closed, right-open interval) ###
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def binary_search_lcro(nums, target)
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# Initialize the left-closed, right-open interval [0, n), with i at the first element and j one past the last
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i, j = 0, nums.length
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# Loop until the search interval is empty (when i = j)
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while i < j
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# Calculate the midpoint index m
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m = (i + j) / 2
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if nums[m] < target
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i = m + 1 # In this case, target must be in [m+1, j) if it exists
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elsif nums[m] > target
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j = m - 1 # In this case, target must be in [i, m) if it exists
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else
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return m # The target element was found; return its index
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end
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end
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-1 # The target element was not found; return -1
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end
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### Driver code ###
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if __FILE__ == $0
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target = 6
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nums = [1, 3, 6, 8, 12, 15, 23, 26, 31, 35]
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# Binary search (closed interval)
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index = binary_search(nums, target)
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puts "Index of target element 6 = #{index}"
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# Binary search (left-closed, right-open interval)
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index = binary_search_lcro(nums, target)
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puts "Index of target element 6 = #{index}"
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end
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