mirror of
https://github.com/krahets/hello-algo.git
synced 2025-11-04 14:18:20 +08:00
docs: add Japanese translate documents (#1812)
* docs: add Japanese documents (`ja/docs`) * docs: add Japanese documents (`ja/codes`) * docs: add Japanese documents * Remove pythontutor blocks in ja/ * Add an empty at the end of each markdown file. * Add the missing figures (use the English version temporarily). * Add index.md for Japanese version. * Add index.html for Japanese version. * Add missing index.assets * Fix backtracking_algorithm.md for Japanese version. * Add avatar_eltociear.jpg. Fix image links on the Japanese landing page. * Add the Japanese banner. --------- Co-authored-by: krahets <krahets@163.com>
This commit is contained in:
Ikko Eltociear Ashimine
committed by
GitHub
GitHub
parent
2487a27036
commit
954c45864b
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"""
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File: iteration.py
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Created Time: 2023-08-24
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Author: krahets (krahets@163.com)
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"""
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def for_loop(n: int) -> int:
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"""forループ"""
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res = 0
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# 1, 2, ..., n-1, n の合計をループ
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for i in range(1, n + 1):
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res += i
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return res
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def while_loop(n: int) -> int:
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"""whileループ"""
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res = 0
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i = 1 # 条件変数を初期化
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# 1, 2, ..., n-1, n の合計をループ
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while i <= n:
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res += i
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i += 1 # 条件変数を更新
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return res
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def while_loop_ii(n: int) -> int:
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"""whileループ（2つの更新）"""
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res = 0
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i = 1 # 条件変数を初期化
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# 1, 4, 10, ... の合計をループ
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while i <= n:
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res += i
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# 条件変数を更新
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i += 1
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i *= 2
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return res
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def nested_for_loop(n: int) -> str:
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"""二重forループ"""
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res = ""
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# i = 1, 2, ..., n-1, n をループ
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for i in range(1, n + 1):
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# j = 1, 2, ..., n-1, n をループ
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for j in range(1, n + 1):
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res += f"({i}, {j}), "
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return res
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"""Driver Code"""
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if __name__ == "__main__":
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n = 5
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res = for_loop(n)
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print(f"\nforループの合計結果 res = {res}")
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res = while_loop(n)
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print(f"\nwhileループの合計結果 res = {res}")
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res = while_loop_ii(n)
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print(f"\nwhileループ（2つの更新）の合計結果 res = {res}")
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res = nested_for_loop(n)
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print(f"\n二重forループの走査結果 {res}")
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@ -0,0 +1,69 @@
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"""
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File: recursion.py
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Created Time: 2023-08-24
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Author: krahets (krahets@163.com)
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"""
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def recur(n: int) -> int:
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"""再帰"""
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# 終了条件
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if n == 1:
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return 1
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# 再帰：再帰呼び出し
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res = recur(n - 1)
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# 復帰：結果を返す
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return n + res
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def for_loop_recur(n: int) -> int:
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"""反復で再帰をシミュレート"""
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# 明示的なスタックを使用してシステムコールスタックをシミュレート
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stack = []
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res = 0
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# 再帰：再帰呼び出し
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for i in range(n, 0, -1):
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# 「スタックへのプッシュ」で「再帰」をシミュレート
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stack.append(i)
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# 復帰：結果を返す
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while stack:
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# 「スタックからのポップ」で「復帰」をシミュレート
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res += stack.pop()
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# res = 1+2+3+...+n
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return res
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def tail_recur(n, res):
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"""末尾再帰"""
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# 終了条件
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if n == 0:
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return res
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# 末尾再帰呼び出し
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return tail_recur(n - 1, res + n)
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def fib(n: int) -> int:
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"""フィボナッチ数列：再帰"""
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# 終了条件 f(1) = 0, f(2) = 1
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if n == 1 or n == 2:
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return n - 1
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# 再帰呼び出し f(n) = f(n-1) + f(n-2)
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res = fib(n - 1) + fib(n - 2)
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# 結果 f(n) を返す
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return res
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"""Driver Code"""
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if __name__ == "__main__":
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n = 5
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res = recur(n)
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print(f"\n再帰関数の合計結果 res = {res}")
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res = for_loop_recur(n)
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print(f"\n反復で再帰をシミュレートする合計結果 res = {res}")
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res = tail_recur(n, 0)
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print(f"\n末尾再帰関数の合計結果 res = {res}")
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res = fib(n)
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print(f"\nフィボナッチ数列の第 {n} 項は {res} です")
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@ -0,0 +1,90 @@
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"""
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File: space_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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import sys
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from pathlib import Path
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sys.path.append(str(Path(__file__).parent.parent))
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from modules import ListNode, TreeNode, print_tree
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def function() -> int:
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"""関数"""
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# 何らかの操作を実行
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return 0
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def constant(n: int):
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"""定数複雑度"""
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# 定数、変数、オブジェクトは O(1) のスペースを占有
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a = 0
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nums = [0] * 10000
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node = ListNode(0)
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# ループ内の変数は O(1) のスペースを占有
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for _ in range(n):
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c = 0
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# ループ内の関数は O(1) のスペースを占有
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for _ in range(n):
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function()
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def linear(n: int):
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"""線形複雑度"""
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# 長さ n のリストは O(n) のスペースを占有
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nums = [0] * n
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# 長さ n のハッシュマップは O(n) のスペースを占有
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hmap = dict[int, str]()
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for i in range(n):
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hmap[i] = str(i)
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def linear_recur(n: int):
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"""線形複雑度（再帰実装）"""
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print("再帰 n =", n)
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if n == 1:
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return
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linear_recur(n - 1)
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def quadratic(n: int):
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"""平方複雑度"""
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# 二次元リストは O(n^2) のスペースを占有
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num_matrix = [[0] * n for _ in range(n)]
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def quadratic_recur(n: int) -> int:
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"""平方複雑度（再帰実装）"""
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if n <= 0:
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return 0
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nums = [0] * n
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print(f"再帰 n = {n} の中で配列の長さ = {len(nums)}")
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return quadratic_recur(n - 1)
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def build_tree(n: int) -> TreeNode | None:
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"""指数複雑度（完全二分木の構築）"""
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if n == 0:
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return None
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root = TreeNode(0)
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root.left = build_tree(n - 1)
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root.right = build_tree(n - 1)
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return root
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"""Driver Code"""
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if __name__ == "__main__":
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n = 5
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# 定数複雑度
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constant(n)
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# 線形複雑度
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linear(n)
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linear_recur(n)
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# 平方複雑度
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quadratic(n)
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quadratic_recur(n)
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# 指数複雑度
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root = build_tree(n)
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print_tree(root)
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@ -0,0 +1,151 @@
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"""
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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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"""定数複雑度"""
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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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"""線形複雑度"""
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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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"""線形複雑度（配列の走査）"""
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count = 0
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# ループ回数は配列の長さに比例する
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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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"""二次複雑度"""
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count = 0
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# ループ回数はデータサイズnの二乗に比例する
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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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"""二次複雑度（バブルソート）"""
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count = 0 # カウンタ
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# 外側のループ: 未ソート範囲は [0, i]
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for i in range(len(nums) - 1, 0, -1):
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# 内側のループ: 未ソート範囲 [0, i] の最大要素を右端にスワップ
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for j in range(i):
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if nums[j] > nums[j + 1]:
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# nums[j] と nums[j + 1] をスワップ
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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 # 要素のスワップは3つの個別操作を含む
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return count
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def exponential(n: int) -> int:
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"""指数複雑度（ループ実装）"""
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count = 0
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base = 1
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# セルは毎回2つに分裂し、1, 2, 4, 8, ..., 2^(n-1) の数列を形成する
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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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"""指数複雑度（再帰実装）"""
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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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"""対数複雑度（ループ実装）"""
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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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"""対数複雑度（再帰実装）"""
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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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"""線形対数複雑度"""
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if n <= 1:
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return 1
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count: int = linear_log_recur(n // 2) + linear_log_recur(n // 2)
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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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"""階乗複雑度（再帰実装）"""
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if n == 0:
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return 1
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count = 0
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# 1つからnに分岐
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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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"""ドライバコード"""
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if __name__ == "__main__":
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# nを変更して、様々な複雑度での操作回数の変化傾向を体験できる
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n = 8
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print("入力データサイズ n =", n)
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count: int = constant(n)
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print("定数複雑度の操作回数 =", count)
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count: int = linear(n)
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print("線形複雑度の操作回数 =", count)
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count: int = array_traversal([0] * n)
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print("線形複雑度（配列の走査）の操作回数 =", count)
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count: int = quadratic(n)
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print("二次複雑度の操作回数 =", count)
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nums = [i for i in range(n, 0, -1)] # [n, n-1, ..., 2, 1]
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count: int = bubble_sort(nums)
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print("二次複雑度（バブルソート）の操作回数 =", count)
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count: int = exponential(n)
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print("指数複雑度（ループ実装）の操作回数 =", count)
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count: int = exp_recur(n)
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print("指数複雑度（再帰実装）の操作回数 =", count)
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count: int = logarithmic(n)
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print("対数複雑度（ループ実装）の操作回数 =", count)
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count: int = log_recur(n)
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print("対数複雑度（再帰実装）の操作回数 =", count)
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count: int = linear_log_recur(n)
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print("線形対数複雑度（再帰実装）の操作回数 =", count)
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count: int = factorial_recur(n)
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print("階乗複雑度（再帰実装）の操作回数 =", count)
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@ -0,0 +1,36 @@
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"""
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File: worst_best_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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import random
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def random_numbers(n: int) -> list[int]:
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"""要素 1, 2, ..., n を含む配列を生成、順序はシャッフル"""
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# 配列 nums = 1, 2, 3, ..., n を生成
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nums = [i for i in range(1, n + 1)]
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# 配列要素をランダムにシャッフル
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random.shuffle(nums)
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return nums
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def find_one(nums: list[int]) -> int:
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"""配列 nums で数値 1 のインデックスを検索"""
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for i in range(len(nums)):
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# 要素 1 が配列の最初にある場合、最良時間計算量 O(1) を達成
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# 要素 1 が配列の最後にある場合、最悪時間計算量 O(n) を達成
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if nums[i] == 1:
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return i
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return -1
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"""Driver Code"""
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if __name__ == "__main__":
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for i in range(10):
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n = 100
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nums: list[int] = random_numbers(n)
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index: int = find_one(nums)
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print("\nシャッフル後の配列 [ 1, 2, ..., n ] =", nums)
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print("数値 1 のインデックス =", index)
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