hello-algo/codes/python/chapter_computational_complexity/time_complexity.py

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"""
File: time_complexity.py
Created Time: 2022-11-25
Author: Krahets (krahets@163.com)
"""
def constant(n: int) -> int:
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""" 常数阶 """
count: int = 0
size: int = 100000
for _ in range(size):
count += 1
return count
def linear(n: int) -> int:
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""" 线性阶 """
count: int = 0
for _ in range(n):
count += 1
return count
def array_traversal(nums: list[int]) -> int:
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""" 线性阶(遍历数组)"""
count: int = 0
# 循环次数与数组长度成正比
for num in nums:
count += 1
return count
def quadratic(n: int) -> int:
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""" 平方阶 """
count: int = 0
# 循环次数与数组长度成平方关系
for i in range(n):
for j in range(n):
count += 1
return count
def bubble_sort(nums: list[int]) -> int:
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""" 平方阶(冒泡排序)"""
count: int = 0 # 计数器
# 外循环:待排序元素数量为 n-1, n-2, ..., 1
for i in range(len(nums) - 1, 0, -1):
# 内循环:冒泡操作
for j in range(i):
if nums[j] > nums[j + 1]:
# 交换 nums[j] 与 nums[j + 1]
tmp: int = nums[j]
nums[j] = nums[j + 1]
nums[j + 1] = tmp
count += 3 # 元素交换包含 3 个单元操作
return count
def exponential(n: int) -> int:
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""" 指数阶(循环实现)"""
count: int = 0
base: int = 1
# cell 每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
for _ in range(n):
for _ in range(base):
count += 1
base *= 2
# count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
return count
def exp_recur(n: int) -> int:
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""" 指数阶(递归实现)"""
if n == 1: return 1
return exp_recur(n - 1) + exp_recur(n - 1) + 1
def logarithmic(n: float) -> int:
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""" 对数阶(循环实现)"""
count: int = 0
while n > 1:
n = n / 2
count += 1
return count
def log_recur(n: float) -> int:
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""" 对数阶(递归实现)"""
if n <= 1: return 0
return log_recur(n / 2) + 1
def linear_log_recur(n: float) -> int:
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""" 线性对数阶 """
if n <= 1: return 1
count: int = linear_log_recur(n // 2) + \
linear_log_recur(n // 2)
for _ in range(n):
count += 1
return count
def factorial_recur(n: int) -> int:
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""" 阶乘阶(递归实现)"""
if n == 0: return 1
count: int = 0
# 从 1 个分裂出 n 个
for _ in range(n):
count += factorial_recur(n - 1)
return count
""" Driver Code """
if __name__ == "__main__":
# 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
n = 8
print("输入数据大小 n =", n)
count: int = constant(n)
print("常数阶的计算操作数量 =", count)
count: int = linear(n)
print("线性阶的计算操作数量 =", count)
count: int = array_traversal([0] * n)
print("线性阶(遍历数组)的计算操作数量 =", count)
count: int = quadratic(n)
print("平方阶的计算操作数量 =", count)
nums: list[int] = [i for i in range(n, 0, -1)] # [n,n-1,...,2,1]
count: int = bubble_sort(nums)
print("平方阶(冒泡排序)的计算操作数量 =", count)
count: int = exponential(n)
print("指数阶(循环实现)的计算操作数量 =", count)
count: int = exp_recur(n)
print("指数阶(递归实现)的计算操作数量 =", count)
count: int = logarithmic(n)
print("对数阶(循环实现)的计算操作数量 =", count)
count: int = log_recur(n)
print("对数阶(递归实现)的计算操作数量 =", count)
count: int = linear_log_recur(n)
print("线性对数阶(递归实现)的计算操作数量 =", count)
count: int = factorial_recur(n)
print("阶乘阶(递归实现)的计算操作数量 =", count)