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* Retranslate the complete English edition from Simplified Chinese * Polish three English explanations after reader-focused review
53 lines
2.0 KiB
Python
53 lines
2.0 KiB
Python
"""
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File: binary_search.py
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Created Time: 2022-11-26
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Author: timi (xisunyy@163.com)
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"""
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def binary_search(nums: list[int], target: int) -> int:
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"""Binary search (closed interval)"""
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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, len(nums) - 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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# Python integers can grow arbitrarily large, subject to available memory, so integer overflow is not a concern
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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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elif 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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return -1 # The target element was not found; return -1
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def binary_search_lcro(nums: list[int], target: int) -> int:
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"""Binary search (left-closed, right-open interval)"""
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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, len(nums)
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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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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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elif nums[m] > target:
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j = m # 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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return -1 # The target element was not found; return -1
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"""Driver code"""
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if __name__ == "__main__":
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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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print("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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print("Index of target element 6 = ", index)
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