mirror of
https://github.com/krahets/hello-algo.git
synced 2026-07-10 06:26:08 +00:00
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 +1,62 @@
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"""
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File: n_queens.py
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Created Time: 2023-04-26
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Author: krahets (krahets@163.com)
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"""
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def backtrack(
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row: int,
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n: int,
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state: list[list[str]],
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res: list[list[list[str]]],
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cols: list[bool],
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diags1: list[bool],
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diags2: list[bool],
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):
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"""回溯演算法：n 皇后"""
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# 當放置完所有行時，記錄解
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if row == n:
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res.append([list(row) for row in state])
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return
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# 走訪所有列
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for col in range(n):
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# 計算該格子對應的主對角線和次對角線
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diag1 = row - col + n - 1
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diag2 = row + col
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# 剪枝：不允許該格子所在列、主對角線、次對角線上存在皇后
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if not cols[col] and not diags1[diag1] and not diags2[diag2]:
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# 嘗試：將皇后放置在該格子
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state[row][col] = "Q"
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cols[col] = diags1[diag1] = diags2[diag2] = True
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# 放置下一行
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backtrack(row + 1, n, state, res, cols, diags1, diags2)
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# 回退：將該格子恢復為空位
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state[row][col] = "#"
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cols[col] = diags1[diag1] = diags2[diag2] = False
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def n_queens(n: int) -> list[list[list[str]]]:
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"""求解 n 皇后"""
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# 初始化 n*n 大小的棋盤，其中 'Q' 代表皇后，'#' 代表空位
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state = [["#" for _ in range(n)] for _ in range(n)]
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cols = [False] * n # 記錄列是否有皇后
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diags1 = [False] * (2 * n - 1) # 記錄主對角線上是否有皇后
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diags2 = [False] * (2 * n - 1) # 記錄次對角線上是否有皇后
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res = []
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backtrack(0, n, state, res, cols, diags1, diags2)
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return res
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"""Driver Code"""
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if __name__ == "__main__":
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n = 4
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res = n_queens(n)
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print(f"輸入棋盤長寬為 {n}")
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print(f"皇后放置方案共有 {len(res)} 種")
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for state in res:
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print("--------------------")
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for row in state:
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print(row)
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@@ -0,0 +1,44 @@
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"""
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File: permutations_i.py
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Created Time: 2023-04-15
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Author: krahets (krahets@163.com)
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"""
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def backtrack(
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state: list[int], choices: list[int], selected: list[bool], res: list[list[int]]
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):
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"""回溯演算法：全排列 I"""
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# 當狀態長度等於元素數量時，記錄解
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if len(state) == len(choices):
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res.append(list(state))
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return
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# 走訪所有選擇
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for i, choice in enumerate(choices):
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# 剪枝：不允許重複選擇元素
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if not selected[i]:
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# 嘗試：做出選擇，更新狀態
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selected[i] = True
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state.append(choice)
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# 進行下一輪選擇
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backtrack(state, choices, selected, res)
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# 回退：撤銷選擇，恢復到之前的狀態
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selected[i] = False
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state.pop()
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def permutations_i(nums: list[int]) -> list[list[int]]:
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"""全排列 I"""
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res = []
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backtrack(state=[], choices=nums, selected=[False] * len(nums), res=res)
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return res
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"""Driver Code"""
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if __name__ == "__main__":
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nums = [1, 2, 3]
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res = permutations_i(nums)
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print(f"輸入陣列 nums = {nums}")
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print(f"所有排列 res = {res}")
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@@ -0,0 +1,46 @@
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"""
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File: permutations_ii.py
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Created Time: 2023-04-15
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Author: krahets (krahets@163.com)
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"""
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def backtrack(
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state: list[int], choices: list[int], selected: list[bool], res: list[list[int]]
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):
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"""回溯演算法：全排列 II"""
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# 當狀態長度等於元素數量時，記錄解
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if len(state) == len(choices):
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res.append(list(state))
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return
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# 走訪所有選擇
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duplicated = set[int]()
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for i, choice in enumerate(choices):
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# 剪枝：不允許重複選擇元素 且 不允許重複選擇相等元素
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if not selected[i] and choice not in duplicated:
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# 嘗試：做出選擇，更新狀態
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duplicated.add(choice) # 記錄選擇過的元素值
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selected[i] = True
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state.append(choice)
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# 進行下一輪選擇
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backtrack(state, choices, selected, res)
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# 回退：撤銷選擇，恢復到之前的狀態
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selected[i] = False
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state.pop()
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def permutations_ii(nums: list[int]) -> list[list[int]]:
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"""全排列 II"""
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res = []
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backtrack(state=[], choices=nums, selected=[False] * len(nums), res=res)
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return res
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"""Driver Code"""
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if __name__ == "__main__":
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nums = [1, 2, 2]
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res = permutations_ii(nums)
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print(f"輸入陣列 nums = {nums}")
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print(f"所有排列 res = {res}")
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@@ -0,0 +1,36 @@
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"""
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File: preorder_traversal_i_compact.py
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Created Time: 2023-04-15
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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 TreeNode, print_tree, list_to_tree
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def pre_order(root: TreeNode):
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"""前序走訪：例題一"""
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if root is None:
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return
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if root.val == 7:
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# 記錄解
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res.append(root)
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pre_order(root.left)
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pre_order(root.right)
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"""Driver Code"""
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if __name__ == "__main__":
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root = list_to_tree([1, 7, 3, 4, 5, 6, 7])
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print("\n初始化二元樹")
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print_tree(root)
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# 前序走訪
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res = list[TreeNode]()
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pre_order(root)
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print("\n輸出所有值為 7 的節點")
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print([node.val for node in res])
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@@ -0,0 +1,42 @@
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"""
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File: preorder_traversal_ii_compact.py
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Created Time: 2023-04-15
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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 TreeNode, print_tree, list_to_tree
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def pre_order(root: TreeNode):
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"""前序走訪：例題二"""
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if root is None:
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return
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# 嘗試
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path.append(root)
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if root.val == 7:
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# 記錄解
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res.append(list(path))
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pre_order(root.left)
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pre_order(root.right)
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# 回退
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path.pop()
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"""Driver Code"""
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if __name__ == "__main__":
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root = list_to_tree([1, 7, 3, 4, 5, 6, 7])
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print("\n初始化二元樹")
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print_tree(root)
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# 前序走訪
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path = list[TreeNode]()
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res = list[list[TreeNode]]()
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pre_order(root)
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print("\n輸出所有根節點到節點 7 的路徑")
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for path in res:
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print([node.val for node in path])
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@@ -0,0 +1,43 @@
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"""
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File: preorder_traversal_iii_compact.py
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Created Time: 2023-04-15
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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 TreeNode, print_tree, list_to_tree
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def pre_order(root: TreeNode):
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"""前序走訪：例題三"""
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# 剪枝
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if root is None or root.val == 3:
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return
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# 嘗試
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path.append(root)
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if root.val == 7:
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# 記錄解
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res.append(list(path))
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pre_order(root.left)
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pre_order(root.right)
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# 回退
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path.pop()
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"""Driver Code"""
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if __name__ == "__main__":
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root = list_to_tree([1, 7, 3, 4, 5, 6, 7])
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print("\n初始化二元樹")
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print_tree(root)
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# 前序走訪
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path = list[TreeNode]()
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res = list[list[TreeNode]]()
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pre_order(root)
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print("\n輸出所有根節點到節點 7 的路徑，路徑中不包含值為 3 的節點")
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for path in res:
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print([node.val for node in path])
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@@ -0,0 +1,71 @@
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"""
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File: preorder_traversal_iii_template.py
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Created Time: 2023-04-15
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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 TreeNode, print_tree, list_to_tree
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def is_solution(state: list[TreeNode]) -> bool:
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"""判斷當前狀態是否為解"""
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return state and state[-1].val == 7
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def record_solution(state: list[TreeNode], res: list[list[TreeNode]]):
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"""記錄解"""
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res.append(list(state))
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def is_valid(state: list[TreeNode], choice: TreeNode) -> bool:
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"""判斷在當前狀態下，該選擇是否合法"""
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return choice is not None and choice.val != 3
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def make_choice(state: list[TreeNode], choice: TreeNode):
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"""更新狀態"""
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state.append(choice)
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def undo_choice(state: list[TreeNode], choice: TreeNode):
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"""恢復狀態"""
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state.pop()
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def backtrack(
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state: list[TreeNode], choices: list[TreeNode], res: list[list[TreeNode]]
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):
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"""回溯演算法：例題三"""
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# 檢查是否為解
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if is_solution(state):
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# 記錄解
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record_solution(state, res)
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# 走訪所有選擇
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for choice in choices:
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# 剪枝：檢查選擇是否合法
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if is_valid(state, choice):
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# 嘗試：做出選擇，更新狀態
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make_choice(state, choice)
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# 進行下一輪選擇
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backtrack(state, [choice.left, choice.right], res)
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# 回退：撤銷選擇，恢復到之前的狀態
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undo_choice(state, choice)
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"""Driver Code"""
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if __name__ == "__main__":
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root = list_to_tree([1, 7, 3, 4, 5, 6, 7])
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print("\n初始化二元樹")
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print_tree(root)
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# 回溯演算法
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res = []
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backtrack(state=[], choices=[root], res=res)
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print("\n輸出所有根節點到節點 7 的路徑，要求路徑中不包含值為 3 的節點")
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for path in res:
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print([node.val for node in path])
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@@ -0,0 +1,48 @@
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"""
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File: subset_sum_i.py
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Created Time: 2023-06-17
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Author: krahets (krahets@163.com)
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"""
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def backtrack(
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state: list[int], target: int, choices: list[int], start: int, res: list[list[int]]
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):
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"""回溯演算法：子集和 I"""
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# 子集和等於 target 時，記錄解
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if target == 0:
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res.append(list(state))
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return
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# 走訪所有選擇
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# 剪枝二：從 start 開始走訪，避免生成重複子集
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for i in range(start, len(choices)):
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# 剪枝一：若子集和超過 target ，則直接結束迴圈
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# 這是因為陣列已排序，後邊元素更大，子集和一定超過 target
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if target - choices[i] < 0:
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break
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# 嘗試：做出選擇，更新 target, start
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state.append(choices[i])
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# 進行下一輪選擇
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backtrack(state, target - choices[i], choices, i, res)
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# 回退：撤銷選擇，恢復到之前的狀態
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state.pop()
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def subset_sum_i(nums: list[int], target: int) -> list[list[int]]:
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"""求解子集和 I"""
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state = [] # 狀態（子集）
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nums.sort() # 對 nums 進行排序
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start = 0 # 走訪起始點
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res = [] # 結果串列（子集串列）
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backtrack(state, target, nums, start, res)
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return res
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"""Driver Code"""
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if __name__ == "__main__":
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nums = [3, 4, 5]
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target = 9
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res = subset_sum_i(nums, target)
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print(f"輸入陣列 nums = {nums}, target = {target}")
|
||||
print(f"所有和等於 {target} 的子集 res = {res}")
|
||||
@@ -0,0 +1,50 @@
|
||||
"""
|
||||
File: subset_sum_i_naive.py
|
||||
Created Time: 2023-06-17
|
||||
Author: krahets (krahets@163.com)
|
||||
"""
|
||||
|
||||
|
||||
def backtrack(
|
||||
state: list[int],
|
||||
target: int,
|
||||
total: int,
|
||||
choices: list[int],
|
||||
res: list[list[int]],
|
||||
):
|
||||
"""回溯演算法：子集和 I"""
|
||||
# 子集和等於 target 時，記錄解
|
||||
if total == target:
|
||||
res.append(list(state))
|
||||
return
|
||||
# 走訪所有選擇
|
||||
for i in range(len(choices)):
|
||||
# 剪枝：若子集和超過 target ，則跳過該選擇
|
||||
if total + choices[i] > target:
|
||||
continue
|
||||
# 嘗試：做出選擇，更新元素和 total
|
||||
state.append(choices[i])
|
||||
# 進行下一輪選擇
|
||||
backtrack(state, target, total + choices[i], choices, res)
|
||||
# 回退：撤銷選擇，恢復到之前的狀態
|
||||
state.pop()
|
||||
|
||||
|
||||
def subset_sum_i_naive(nums: list[int], target: int) -> list[list[int]]:
|
||||
"""求解子集和 I（包含重複子集）"""
|
||||
state = [] # 狀態（子集）
|
||||
total = 0 # 子集和
|
||||
res = [] # 結果串列（子集串列）
|
||||
backtrack(state, target, total, nums, res)
|
||||
return res
|
||||
|
||||
|
||||
"""Driver Code"""
|
||||
if __name__ == "__main__":
|
||||
nums = [3, 4, 5]
|
||||
target = 9
|
||||
res = subset_sum_i_naive(nums, target)
|
||||
|
||||
print(f"輸入陣列 nums = {nums}, target = {target}")
|
||||
print(f"所有和等於 {target} 的子集 res = {res}")
|
||||
print(f"請注意，該方法輸出的結果包含重複集合")
|
||||
@@ -0,0 +1,52 @@
|
||||
"""
|
||||
File: subset_sum_ii.py
|
||||
Created Time: 2023-06-17
|
||||
Author: krahets (krahets@163.com)
|
||||
"""
|
||||
|
||||
|
||||
def backtrack(
|
||||
state: list[int], target: int, choices: list[int], start: int, res: list[list[int]]
|
||||
):
|
||||
"""回溯演算法：子集和 II"""
|
||||
# 子集和等於 target 時，記錄解
|
||||
if target == 0:
|
||||
res.append(list(state))
|
||||
return
|
||||
# 走訪所有選擇
|
||||
# 剪枝二：從 start 開始走訪，避免生成重複子集
|
||||
# 剪枝三：從 start 開始走訪，避免重複選擇同一元素
|
||||
for i in range(start, len(choices)):
|
||||
# 剪枝一：若子集和超過 target ，則直接結束迴圈
|
||||
# 這是因為陣列已排序，後邊元素更大，子集和一定超過 target
|
||||
if target - choices[i] < 0:
|
||||
break
|
||||
# 剪枝四：如果該元素與左邊元素相等，說明該搜尋分支重複，直接跳過
|
||||
if i > start and choices[i] == choices[i - 1]:
|
||||
continue
|
||||
# 嘗試：做出選擇，更新 target, start
|
||||
state.append(choices[i])
|
||||
# 進行下一輪選擇
|
||||
backtrack(state, target - choices[i], choices, i + 1, res)
|
||||
# 回退：撤銷選擇，恢復到之前的狀態
|
||||
state.pop()
|
||||
|
||||
|
||||
def subset_sum_ii(nums: list[int], target: int) -> list[list[int]]:
|
||||
"""求解子集和 II"""
|
||||
state = [] # 狀態（子集）
|
||||
nums.sort() # 對 nums 進行排序
|
||||
start = 0 # 走訪起始點
|
||||
res = [] # 結果串列（子集串列）
|
||||
backtrack(state, target, nums, start, res)
|
||||
return res
|
||||
|
||||
|
||||
"""Driver Code"""
|
||||
if __name__ == "__main__":
|
||||
nums = [4, 4, 5]
|
||||
target = 9
|
||||
res = subset_sum_ii(nums, target)
|
||||
|
||||
print(f"輸入陣列 nums = {nums}, target = {target}")
|
||||
print(f"所有和等於 {target} 的子集 res = {res}")
|
||||
Reference in New Issue
Block a user