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55 lines
1.8 KiB
Python
55 lines
1.8 KiB
Python
"""
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File: binary_search_insertion.py
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Created Time: 2023-08-04
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Author: krahets (krahets@163.com)
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"""
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def binary_search_insertion_simple(nums: list[int], target: int) -> int:
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"""二分查找插入点（无重复元素）"""
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i, j = 0, len(nums) - 1 # 初始化双闭区间 [0, n-1]
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while i <= j:
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m = (i + j) // 2 # 计算中点索引 m
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if nums[m] < target:
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i = m + 1 # target 在区间 [m+1, j] 中
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elif nums[m] > target:
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j = m - 1 # target 在区间 [i, m-1] 中
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else:
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return m # 找到 target ，返回插入点 m
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# 未找到 target ，返回插入点 i
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return i
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def binary_search_insertion(nums: list[int], target: int) -> int:
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"""二分查找插入点（存在重复元素）"""
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i, j = 0, len(nums) - 1 # 初始化双闭区间 [0, n-1]
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while i <= j:
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m = (i + j) // 2 # 计算中点索引 m
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if nums[m] < target:
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i = m + 1 # target 在区间 [m+1, j] 中
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elif nums[m] > target:
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j = m - 1 # target 在区间 [i, m-1] 中
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else:
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j = m - 1 # 首个小于 target 的元素在区间 [i, m-1] 中
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# 返回插入点 i
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return i
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"""Driver Code"""
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if __name__ == "__main__":
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# 无重复元素的数组
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nums = [1, 3, 6, 8, 12, 15, 23, 26, 31, 35]
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print(f"\n数组 nums = {nums}")
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# 二分查找插入点
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for target in [6, 9]:
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index = binary_search_insertion_simple(nums, target)
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print(f"元素 {target} 的插入点的索引为 {index}")
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# 包含重复元素的数组
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nums = [1, 3, 6, 6, 6, 6, 6, 10, 12, 15]
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print(f"\n数组 nums = {nums}")
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# 二分查找插入点
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for target in [2, 6, 20]:
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index = binary_search_insertion(nums, target)
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print(f"元素 {target} 的插入点的索引为 {index}")
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