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1f784dadb0
divide and conquer.
40 lines
1 KiB
Python
40 lines
1 KiB
Python
"""
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File: binary_search_recur.py
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Created Time: 2023-07-17
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Author: krahets (krahets@163.com)
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"""
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def dfs(nums: list[int], target: int, i: int, j: int) -> int:
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"""二分查找:问题 f(i, j)"""
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# 若区间为空,代表无目标元素,则返回 -1
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if i > j:
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return -1
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# 计算中点索引 m
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m = (i + j) // 2
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if nums[m] < target:
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# 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j)
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elif nums[m] > target:
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# 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1)
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else:
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# 找到目标元素,返回其索引
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return m
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def binary_search(nums: list[int], target: int) -> int:
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"""二分查找"""
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n = len(nums)
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# 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 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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# 二分查找(双闭区间)
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index: int = binary_search(nums, target)
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print("目标元素 6 的索引 = ", index)
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