mirror of
https://github.com/krahets/hello-algo.git
synced 2025-11-01 20:12:07 +08:00
feat: Traditional Chinese version (#1163)
* First commit * Update mkdocs.yml * Translate all the docs to traditional Chinese * Translate the code files. * Translate the docker file * Fix mkdocs.yml * Translate all the figures from SC to TC * 二叉搜尋樹 -> 二元搜尋樹 * Update terminology. * Update terminology * 构造函数/构造方法 -> 建構子 异或 -> 互斥或 * 擴充套件 -> 擴展 * constant - 常量 - 常數 * 類 -> 類別 * AVL -> AVL 樹 * 數組 -> 陣列 * 係統 -> 系統 斐波那契數列 -> 費波那契數列 運算元量 -> 運算量 引數 -> 參數 * 聯絡 -> 關聯 * 麵試 -> 面試 * 面向物件 -> 物件導向 歸併排序 -> 合併排序 范式 -> 範式 * Fix 算法 -> 演算法 * 錶示 -> 表示 反碼 -> 一補數 補碼 -> 二補數 列列尾部 -> 佇列尾部 區域性性 -> 區域性 一摞 -> 一疊 * Synchronize with main branch * 賬號 -> 帳號 推匯 -> 推導 * Sync with main branch * First commit * Update mkdocs.yml * Translate all the docs to traditional Chinese * Translate the code files. * Translate the docker file * Fix mkdocs.yml * Translate all the figures from SC to TC * 二叉搜尋樹 -> 二元搜尋樹 * Update terminology * 构造函数/构造方法 -> 建構子 异或 -> 互斥或 * 擴充套件 -> 擴展 * constant - 常量 - 常數 * 類 -> 類別 * AVL -> AVL 樹 * 數組 -> 陣列 * 係統 -> 系統 斐波那契數列 -> 費波那契數列 運算元量 -> 運算量 引數 -> 參數 * 聯絡 -> 關聯 * 麵試 -> 面試 * 面向物件 -> 物件導向 歸併排序 -> 合併排序 范式 -> 範式 * Fix 算法 -> 演算法 * 錶示 -> 表示 反碼 -> 一補數 補碼 -> 二補數 列列尾部 -> 佇列尾部 區域性性 -> 區域性 一摞 -> 一疊 * Synchronize with main branch * 賬號 -> 帳號 推匯 -> 推導 * Sync with main branch * Update terminology.md * 操作数量（num. of operations）-> 操作數量 * 字首和->前綴和 * Update figures * 歸 -> 迴 記憶體洩漏 -> 記憶體流失 * Fix the bug of the file filter * 支援 -> 支持 Add zh-Hant/README.md * Add the zh-Hant chapter covers. Bug fixes. * 外掛 -> 擴充功能 * Add the landing page for zh-Hant version * Unify the font of the chapter covers for the zh, en, and zh-Hant version * Move zh-Hant/ to zh-hant/ * Translate terminology.md to traditional Chinese
This commit is contained in:
Yudong Jin
@ -0,0 +1,65 @@
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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 迴圈"""
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res = 0
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# 迴圈求和 1, 2, ..., n-1, n
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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 迴圈"""
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res = 0
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i = 1 # 初始化條件變數
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# 迴圈求和 1, 2, ..., n-1, n
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while i <= n:
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res += i
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i += 1 # 更新條件變數
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return res
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def while_loop_ii(n: int) -> int:
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"""while 迴圈（兩次更新）"""
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res = 0
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i = 1 # 初始化條件變數
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# 迴圈求和 1, 4, 10, ...
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while i <= n:
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res += i
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# 更新條件變數
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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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"""雙層 for 迴圈"""
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res = ""
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# 迴圈 i = 1, 2, ..., n-1, n
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for i in range(1, n + 1):
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# 迴圈 j = 1, 2, ..., n-1, n
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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 迴圈的求和結果 res = {res}")
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res = while_loop(n)
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print(f"\nwhile 迴圈的求和結果 res = {res}")
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res = while_loop_ii(n)
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print(f"\nwhile 迴圈（兩次更新）求和結果 res = {res}")
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res = nested_for_loop(n)
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print(f"\n雙層 for 迴圈的走訪結果 {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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"""遞迴"""
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# 終止條件
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if n == 1:
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return 1
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# 遞：遞迴呼叫
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res = recur(n - 1)
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# 迴：返回結果
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return n + res
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def for_loop_recur(n: int) -> int:
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"""使用迭代模擬遞迴"""
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# 使用一個顯式的堆疊來模擬系統呼叫堆疊
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stack = []
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res = 0
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# 遞：遞迴呼叫
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for i in range(n, 0, -1):
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# 透過“入堆疊操作”模擬“遞”
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stack.append(i)
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# 迴：返回結果
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while stack:
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# 透過“出堆疊操作”模擬“迴”
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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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"""尾遞迴"""
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# 終止條件
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if n == 0:
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return res
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# 尾遞迴呼叫
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return tail_recur(n - 1, res + n)
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def fib(n: int) -> int:
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"""費波那契數列：遞迴"""
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# 終止條件 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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# 遞迴呼叫 f(n) = f(n-1) + f(n-2)
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res = fib(n - 1) + fib(n - 2)
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# 返回結果 f(n)
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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遞迴函式的求和結果 res = {res}")
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res = for_loop_recur(n)
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print(f"\n使用迭代模擬遞迴求和結果 res = {res}")
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res = tail_recur(n, 0)
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print(f"\n尾遞迴函式的求和結果 res = {res}")
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res = fib(n)
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print(f"\n費波那契數列的第 {n} 項為 {res}")
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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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"""函式"""
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# 執行某些操作
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return 0
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def constant(n: int):
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"""常數階"""
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# 常數、變數、物件佔用 O(1) 空間
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a = 0
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nums = [0] * 10000
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node = ListNode(0)
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# 迴圈中的變數佔用 O(1) 空間
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for _ in range(n):
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c = 0
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# 迴圈中的函式佔用 O(1) 空間
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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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"""線性階"""
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# 長度為 n 的串列佔用 O(n) 空間
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nums = [0] * n
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# 長度為 n 的雜湊表佔用 O(n) 空間
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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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"""線性階（遞迴實現）"""
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print("遞迴 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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"""平方階"""
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# 二維串列佔用 O(n^2) 空間
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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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"""平方階（遞迴實現）"""
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if n <= 0:
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return 0
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# 陣列 nums 長度為 n, n-1, ..., 2, 1
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nums = [0] * n
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return quadratic_recur(n - 1)
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def build_tree(n: int) -> TreeNode | None:
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"""指數階（建立滿二元樹）"""
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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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# 常數階
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constant(n)
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# 線性階
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linear(n)
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linear_recur(n)
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# 平方階
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quadratic(n)
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quadratic_recur(n)
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# 指數階
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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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"""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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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: int = constant(n)
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print("常數階的操作數量 =", count)
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count: int = linear(n)
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print("線性階的操作數量 =", count)
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count: int = 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: int = 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: int = 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: int = 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: int = 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: int = 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: int = 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: int = 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: int = 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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@ -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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|
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import random
|
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|
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|
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def random_numbers(n: int) -> list[int]:
|
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"""生成一個陣列，元素為: 1, 2, ..., n ，順序被打亂"""
|
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# 生成陣列 nums =: 1, 2, 3, ..., n
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nums = [i for i in range(1, n + 1)]
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# 隨機打亂陣列元素
|
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random.shuffle(nums)
|
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return nums
|
||||
|
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|
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def find_one(nums: list[int]) -> int:
|
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"""查詢陣列 nums 中數字 1 所在索引"""
|
||||
for i in range(len(nums)):
|
||||
# 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
||||
# 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
||||
if nums[i] == 1:
|
||||
return i
|
||||
return -1
|
||||
|
||||
|
||||
"""Driver Code"""
|
||||
if __name__ == "__main__":
|
||||
for i in range(10):
|
||||
n = 100
|
||||
nums: list[int] = random_numbers(n)
|
||||
index: int = find_one(nums)
|
||||
print("\n陣列 [ 1, 2, ..., n ] 被打亂後 =", nums)
|
||||
print("數字 1 的索引為", index)
|
||||
Reference in New Issue
Block a user