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* preorder, inorder, postorder -> pre-order, in-order, post-order * Bug fixes * Bug fixes * Update what_is_dsa.md * Sync zh and zh-hant versions * Sync zh and zh-hant versions. * Update performance_evaluation.md and time_complexity.md * Add @khoaxuantu to the landing page. * Sync zh and zh-hant versions * Add @ khoaxuantu to the landing page of zh-hant and en versions.
153 lines
3.8 KiB
Python
153 lines
3.8 KiB
Python
"""
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File: time_complexity.py
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Created Time: 2022-11-25
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Author: krahets (krahets@163.com)
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"""
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def constant(n: int) -> int:
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"""常数阶"""
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count = 0
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size = 100000
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for _ in range(size):
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count += 1
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return count
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def linear(n: int) -> int:
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"""线性阶"""
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count = 0
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for _ in range(n):
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count += 1
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return count
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def array_traversal(nums: list[int]) -> int:
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"""线性阶（遍历数组）"""
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count = 0
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# 循环次数与数组长度成正比
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for num in nums:
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count += 1
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return count
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def quadratic(n: int) -> int:
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"""平方阶"""
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count = 0
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# 循环次数与数据大小 n 成平方关系
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for i in range(n):
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for j in range(n):
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count += 1
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return count
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def bubble_sort(nums: list[int]) -> int:
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"""平方阶（冒泡排序）"""
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count = 0 # 计数器
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# 外循环：未排序区间为 [0, i]
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for i in range(len(nums) - 1, 0, -1):
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# 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
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for j in range(i):
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if nums[j] > nums[j + 1]:
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# 交换 nums[j] 与 nums[j + 1]
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tmp: int = nums[j]
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nums[j] = nums[j + 1]
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nums[j + 1] = tmp
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count += 3 # 元素交换包含 3 个单元操作
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return count
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def exponential(n: int) -> int:
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"""指数阶（循环实现）"""
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count = 0
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base = 1
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# 细胞每轮一分为二，形成数列 1, 2, 4, 8, ..., 2^(n-1)
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for _ in range(n):
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for _ in range(base):
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count += 1
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base *= 2
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# count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count
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def exp_recur(n: int) -> int:
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"""指数阶（递归实现）"""
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if n == 1:
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return 1
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return exp_recur(n - 1) + exp_recur(n - 1) + 1
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def logarithmic(n: int) -> int:
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"""对数阶（循环实现）"""
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count = 0
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while n > 1:
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n = n / 2
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count += 1
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return count
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def log_recur(n: int) -> int:
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"""对数阶（递归实现）"""
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if n <= 1:
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return 0
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return log_recur(n / 2) + 1
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def linear_log_recur(n: int) -> int:
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"""线性对数阶"""
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if n <= 1:
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return 1
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# 一分为二，子问题的规模减小一半
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count = linear_log_recur(n // 2) + linear_log_recur(n // 2)
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# 当前子问题包含 n 个操作
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for _ in range(n):
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count += 1
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return count
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def factorial_recur(n: int) -> int:
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"""阶乘阶（递归实现）"""
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if n == 0:
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return 1
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count = 0
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# 从 1 个分裂出 n 个
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for _ in range(n):
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count += factorial_recur(n - 1)
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return count
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"""Driver Code"""
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if __name__ == "__main__":
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# 可以修改 n 运行，体会一下各种复杂度的操作数量变化趋势
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n = 8
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print("输入数据大小 n =", n)
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count = constant(n)
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print("常数阶的操作数量 =", count)
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count = linear(n)
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print("线性阶的操作数量 =", count)
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count = array_traversal([0] * n)
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print("线性阶（遍历数组）的操作数量 =", count)
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count = quadratic(n)
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print("平方阶的操作数量 =", count)
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nums = [i for i in range(n, 0, -1)] # [n, n-1, ..., 2, 1]
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count = bubble_sort(nums)
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print("平方阶（冒泡排序）的操作数量 =", count)
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count = exponential(n)
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print("指数阶（循环实现）的操作数量 =", count)
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count = exp_recur(n)
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print("指数阶（递归实现）的操作数量 =", count)
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count = logarithmic(n)
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print("对数阶（循环实现）的操作数量 =", count)
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count = log_recur(n)
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print("对数阶（递归实现）的操作数量 =", count)
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count = linear_log_recur(n)
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print("线性对数阶（递归实现）的操作数量 =", count)
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count = factorial_recur(n)
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print("阶乘阶（递归实现）的操作数量 =", count)
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