117 lines
3.9 KiB
Python
117 lines
3.9 KiB
Python
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"""
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使用二分查找的前提是，查找的序列是已经排序过的，时间复杂度为 O(nlog2n)
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1. 使用两个指针low, high，分别指向第一个元素和最后一个元素
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2. 取low 和 high 之间的中间值，并将此值与要查找的值比较
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(2.1) 若小于查找值，向右缩小范围，low 移向中间，high 不变
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(2.2) 若大于查找值，则向左缩小范围，high 移向中间，low 不变
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3. 重复第2步
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"""
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def binary_search(sequence: list[int], value):
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low = 0
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high = len(sequence) - 1
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mid = (low + high) // 2
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while low != high:
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if sequence[mid] > value:
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high = mid - 1
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elif sequence[mid] < value:
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low = mid + 1
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else:
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return mid
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mid = (low + high) // 2
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if sequence[low] == value:
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return low
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else:
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return False
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def binary_search_recur(sequence, value, low, high):
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if sequence[(low + high) // 2] == value:
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return (low + high) // 2
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if low == high:
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return low if sequence[low] == value else False
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if sequence[(low + high) // 2] > value:
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return binary_search_recur(sequence, value, low, (low + high) // 2 - 1)
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if sequence[(low + high) // 2] < value:
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return binary_search_recur(sequence, value, (low + high) // 2 + 1, high)
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"""
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使用内插查找：时间复杂度为 O(loglog2n)
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内插查找是一种二分查找的变形，适合在排序数据中进行查找。
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内插查找不是像二分查找算法中那样直接使用中值来定界，而是通过插值算法找到上下
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界。
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类似于计算一条直线的函数: y=kx。在已排序的序列中，取两点 [x0, y0], [x1, y1]，即可计算出这两
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点之间的任何值：(x-x0)/(y-y0) = (x0-x1)/(y0-y1) x=(y-y0)(x1-x0)/(y1-y0) + x0 => 直线斜率计算式
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因此对于一个序列，下标可作为x，值为y
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1. 取第一个元素的下表和值为 x0, y0，最后的元素的下标和值为x1, y1
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2. 将要查找的值作为 y, 通过斜率公式计算对应的 x
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3. 取下标为 x 的值，若值大于 y，将这个下标和值作为上界，反之作为下界
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4. 重复2，3
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"""
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def interpolation_search(sequence, value):
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low = 0
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high = len(sequence) - 1
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while low < high:
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x = (value - sequence[low]) * (high - low) // (sequence[high] - sequence[low]) + low
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if sequence[low] > value:
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return False
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if sequence[x] > value:
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high = x - 1
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elif sequence[x] < value:
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low = x + 1
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else:
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return x
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if sequence[low] == value:
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return low
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else:
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return False
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"""
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指数查找：它划分中值的方法不是使用平均或插值而是用指数函数来估计，这样可以快速找到上界
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该算法适合已排序且无边界的数据。
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算法查找过程中不断比较 2^0, 2^1, 2^2, 2^k 位置上的值和目标值的关系，进而确定搜索区域，之后在
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该区域内使用二分查找算法查找
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假设要在 [2,3,4,6,7,8,10,13,15,19,20,22,23,24,28] 这个 15 个元素已排序集合中查找 22，
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那么首先查看 2
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0 = 1 位置上的数字是否超过 22，得到 3 < 22，所以继续查找 2^1, 2^2, 2^3 位置
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处元素，发现对应的值 4, 7, 15 均小于 22。继续查看 16 = 24 处的值，可是 16 大于集合元
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素个数，超出范围了，所以查找上界就是最后一个索引 14。
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注意下界是 high 的一半，能找到一个上界，那么说明前一次访问处也就是 2^(n-1)
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一定小于待查找的值，作为下界是合理的
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"""
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def exponential_search(sequence: list[int], value):
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size = len(sequence)
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# 由于下界取上界的一半，所以 high 从 1 开始
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high = 1
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while high < size and sequence[high] < value:
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high <<= 1
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low = high >> 1
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res = binary_search(sequence[low: high + 1], value)
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return res + low if res else res
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sequence = [1, 4, 6, 10, 14, 18, 24, 39, 50]
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num1 = 10
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exponential_search(sequence, num1)
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