You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
154 lines
3.8 KiB
154 lines
3.8 KiB
"""
|
|
File: time_complexity.py
|
|
Created Time: 2022-11-25
|
|
Author: krahets (krahets@163.com)
|
|
"""
|
|
|
|
|
|
def constant(n: int) -> int:
|
|
"""常數階"""
|
|
count = 0
|
|
size = 100000
|
|
for _ in range(size):
|
|
count += 1
|
|
return count
|
|
|
|
|
|
def linear(n: int) -> int:
|
|
"""線性階"""
|
|
count = 0
|
|
for _ in range(n):
|
|
count += 1
|
|
return count
|
|
|
|
|
|
def array_traversal(nums: list[int]) -> int:
|
|
"""線性階（走訪陣列）"""
|
|
count = 0
|
|
# 迴圈次數與陣列長度成正比
|
|
for num in nums:
|
|
count += 1
|
|
return count
|
|
|
|
|
|
def quadratic(n: int) -> int:
|
|
"""平方階"""
|
|
count = 0
|
|
# 迴圈次數與資料大小 n 成平方關係
|
|
for i in range(n):
|
|
for j in range(n):
|
|
count += 1
|
|
return count
|
|
|
|
|
|
def bubble_sort(nums: list[int]) -> int:
|
|
"""平方階（泡沫排序）"""
|
|
count = 0 # 計數器
|
|
# 外迴圈：未排序區間為 [0, i]
|
|
for i in range(len(nums) - 1, 0, -1):
|
|
# 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
for j in range(i):
|
|
if nums[j] > nums[j + 1]:
|
|
# 交換 nums[j] 與 nums[j + 1]
|
|
tmp: int = nums[j]
|
|
nums[j] = nums[j + 1]
|
|
nums[j + 1] = tmp
|
|
count += 3 # 元素交換包含 3 個單元操作
|
|
return count
|
|
|
|
|
|
def exponential(n: int) -> int:
|
|
"""指數階（迴圈實現）"""
|
|
count = 0
|
|
base = 1
|
|
# 細胞每輪一分為二，形成數列 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:
|
|
"""指數階（遞迴實現）"""
|
|
if n == 1:
|
|
return 1
|
|
return exp_recur(n - 1) + exp_recur(n - 1) + 1
|
|
|
|
|
|
def logarithmic(n: int) -> int:
|
|
"""對數階（迴圈實現）"""
|
|
count = 0
|
|
while n > 1:
|
|
n = n / 2
|
|
count += 1
|
|
return count
|
|
|
|
|
|
def log_recur(n: int) -> int:
|
|
"""對數階（遞迴實現）"""
|
|
if n <= 1:
|
|
return 0
|
|
return log_recur(n / 2) + 1
|
|
|
|
|
|
def linear_log_recur(n: int) -> int:
|
|
"""線性對數階"""
|
|
if n <= 1:
|
|
return 1
|
|
# 一分為二，子問題的規模減小一半
|
|
count = linear_log_recur(n // 2) + linear_log_recur(n // 2)
|
|
# 當前子問題包含 n 個操作
|
|
for _ in range(n):
|
|
count += 1
|
|
return count
|
|
|
|
|
|
def factorial_recur(n: int) -> int:
|
|
"""階乘階（遞迴實現）"""
|
|
if n == 0:
|
|
return 1
|
|
count = 0
|
|
# 從 1 個分裂出 n 個
|
|
for _ in range(n):
|
|
count += factorial_recur(n - 1)
|
|
return count
|
|
|
|
|
|
"""Driver Code"""
|
|
if __name__ == "__main__":
|
|
# 可以修改 n 執行，體會一下各種複雜度的操作數量變化趨勢
|
|
n = 8
|
|
print("輸入資料大小 n =", n)
|
|
|
|
count = constant(n)
|
|
print("常數階的操作數量 =", count)
|
|
|
|
count = linear(n)
|
|
print("線性階的操作數量 =", count)
|
|
count = array_traversal([0] * n)
|
|
print("線性階（走訪陣列）的操作數量 =", count)
|
|
|
|
count = quadratic(n)
|
|
print("平方階的操作數量 =", count)
|
|
nums = [i for i in range(n, 0, -1)] # [n, n-1, ..., 2, 1]
|
|
count = bubble_sort(nums)
|
|
print("平方階（泡沫排序）的操作數量 =", count)
|
|
|
|
count = exponential(n)
|
|
print("指數階（迴圈實現）的操作數量 =", count)
|
|
count = exp_recur(n)
|
|
print("指數階（遞迴實現）的操作數量 =", count)
|
|
|
|
count = logarithmic(n)
|
|
print("對數階（迴圈實現）的操作數量 =", count)
|
|
count = log_recur(n)
|
|
print("對數階（遞迴實現）的操作數量 =", count)
|
|
|
|
count = linear_log_recur(n)
|
|
print("線性對數階（遞迴實現）的操作數量 =", count)
|
|
|
|
count = factorial_recur(n)
|
|
print("階乘階（遞迴實現）的操作數量 =", count)
|