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3f4220de81
* 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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"""常數階"""
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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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"""線性階"""
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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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"""線性階（走訪陣列）"""
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count = 0
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# 迴圈次數與陣列長度成正比
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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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"""平方階"""
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count = 0
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# 迴圈次數與資料大小 n 成平方關係
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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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"""平方階（泡沫排序）"""
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count = 0 # 計數器
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# 外迴圈：未排序區間為 [0, i]
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for i in range(len(nums) - 1, 0, -1):
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# 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
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for j in range(i):
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if nums[j] > nums[j + 1]:
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# 交換 nums[j] 與 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 # 元素交換包含 3 個單元操作
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return count
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def exponential(n: int) -> int:
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"""指數階（迴圈實現）"""
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count = 0
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base = 1
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# 細胞每輪一分為二，形成數列 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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"""指數階（遞迴實現）"""
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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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"""對數階（迴圈實現）"""
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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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"""對數階（遞迴實現）"""
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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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"""線性對數階"""
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if n <= 1:
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return 1
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# 一分為二，子問題的規模減小一半
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count = linear_log_recur(n // 2) + linear_log_recur(n // 2)
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# 當前子問題包含 n 個操作
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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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"""階乘階（遞迴實現）"""
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if n == 0:
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return 1
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count = 0
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# 從 1 個分裂出 n 個
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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 執行，體會一下各種複雜度的操作數量變化趨勢
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n = 8
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print("輸入資料大小 n =", n)
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count = constant(n)
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print("常數階的操作數量 =", count)
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count = linear(n)
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print("線性階的操作數量 =", count)
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count = array_traversal([0] * n)
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print("線性階（走訪陣列）的操作數量 =", count)
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count = quadratic(n)
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print("平方階的操作數量 =", 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("平方階（泡沫排序）的操作數量 =", count)
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count = exponential(n)
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print("指數階（迴圈實現）的操作數量 =", count)
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count = exp_recur(n)
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print("指數階（遞迴實現）的操作數量 =", count)
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count = logarithmic(n)
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print("對數階（迴圈實現）的操作數量 =", count)
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count = log_recur(n)
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print("對數階（遞迴實現）的操作數量 =", count)
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count = linear_log_recur(n)
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print("線性對數階（遞迴實現）的操作數量 =", count)
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count = factorial_recur(n)
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print("階乘階（遞迴實現）的操作數量 =", count)
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