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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:
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"""
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File: climbing_stairs_backtrack.py
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Created Time: 2023-06-30
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Author: krahets (krahets@163.com)
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"""
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def backtrack(choices: list[int], state: int, n: int, res: list[int]) -> int:
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"""回溯"""
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# 當爬到第 n 階時,方案數量加 1
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if state == n:
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res[0] += 1
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# 走訪所有選擇
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for choice in choices:
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# 剪枝:不允許越過第 n 階
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if state + choice > n:
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continue
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# 嘗試:做出選擇,更新狀態
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backtrack(choices, state + choice, n, res)
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# 回退
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def climbing_stairs_backtrack(n: int) -> int:
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"""爬樓梯:回溯"""
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choices = [1, 2] # 可選擇向上爬 1 階或 2 階
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state = 0 # 從第 0 階開始爬
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res = [0] # 使用 res[0] 記錄方案數量
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backtrack(choices, state, n, res)
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return res[0]
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"""Driver Code"""
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if __name__ == "__main__":
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n = 9
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res = climbing_stairs_backtrack(n)
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print(f"爬 {n} 階樓梯共有 {res} 種方案")
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@ -0,0 +1,29 @@
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"""
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File: climbing_stairs_constraint_dp.py
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Created Time: 2023-06-30
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Author: krahets (krahets@163.com)
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"""
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def climbing_stairs_constraint_dp(n: int) -> int:
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"""帶約束爬樓梯:動態規劃"""
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if n == 1 or n == 2:
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return 1
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# 初始化 dp 表,用於儲存子問題的解
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dp = [[0] * 3 for _ in range(n + 1)]
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# 初始狀態:預設最小子問題的解
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dp[1][1], dp[1][2] = 1, 0
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dp[2][1], dp[2][2] = 0, 1
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# 狀態轉移:從較小子問題逐步求解較大子問題
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for i in range(3, n + 1):
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dp[i][1] = dp[i - 1][2]
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dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
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return dp[n][1] + dp[n][2]
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"""Driver Code"""
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if __name__ == "__main__":
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n = 9
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res = climbing_stairs_constraint_dp(n)
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print(f"爬 {n} 階樓梯共有 {res} 種方案")
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@ -0,0 +1,28 @@
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"""
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File: climbing_stairs_dfs.py
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Created Time: 2023-06-30
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Author: krahets (krahets@163.com)
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"""
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def dfs(i: int) -> int:
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"""搜尋"""
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# 已知 dp[1] 和 dp[2] ,返回之
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if i == 1 or i == 2:
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return i
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# dp[i] = dp[i-1] + dp[i-2]
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count = dfs(i - 1) + dfs(i - 2)
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return count
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def climbing_stairs_dfs(n: int) -> int:
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"""爬樓梯:搜尋"""
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return dfs(n)
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"""Driver Code"""
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if __name__ == "__main__":
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n = 9
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res = climbing_stairs_dfs(n)
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print(f"爬 {n} 階樓梯共有 {res} 種方案")
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@ -0,0 +1,35 @@
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"""
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File: climbing_stairs_dfs_mem.py
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Created Time: 2023-06-30
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Author: krahets (krahets@163.com)
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"""
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def dfs(i: int, mem: list[int]) -> int:
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"""記憶化搜尋"""
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# 已知 dp[1] 和 dp[2] ,返回之
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if i == 1 or i == 2:
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return i
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# 若存在記錄 dp[i] ,則直接返回之
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if mem[i] != -1:
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return mem[i]
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# dp[i] = dp[i-1] + dp[i-2]
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count = dfs(i - 1, mem) + dfs(i - 2, mem)
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# 記錄 dp[i]
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mem[i] = count
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return count
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def climbing_stairs_dfs_mem(n: int) -> int:
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"""爬樓梯:記憶化搜尋"""
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# mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
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mem = [-1] * (n + 1)
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return dfs(n, mem)
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"""Driver Code"""
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if __name__ == "__main__":
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n = 9
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res = climbing_stairs_dfs_mem(n)
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print(f"爬 {n} 階樓梯共有 {res} 種方案")
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@ -0,0 +1,40 @@
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"""
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File: climbing_stairs_dp.py
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Created Time: 2023-06-30
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Author: krahets (krahets@163.com)
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"""
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def climbing_stairs_dp(n: int) -> int:
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"""爬樓梯:動態規劃"""
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if n == 1 or n == 2:
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return n
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# 初始化 dp 表,用於儲存子問題的解
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dp = [0] * (n + 1)
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# 初始狀態:預設最小子問題的解
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dp[1], dp[2] = 1, 2
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# 狀態轉移:從較小子問題逐步求解較大子問題
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for i in range(3, n + 1):
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dp[i] = dp[i - 1] + dp[i - 2]
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return dp[n]
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def climbing_stairs_dp_comp(n: int) -> int:
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"""爬樓梯:空間最佳化後的動態規劃"""
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if n == 1 or n == 2:
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return n
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a, b = 1, 2
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for _ in range(3, n + 1):
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a, b = b, a + b
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return b
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"""Driver Code"""
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if __name__ == "__main__":
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n = 9
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res = climbing_stairs_dp(n)
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print(f"爬 {n} 階樓梯共有 {res} 種方案")
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res = climbing_stairs_dp_comp(n)
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print(f"爬 {n} 階樓梯共有 {res} 種方案")
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@ -0,0 +1,60 @@
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"""
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File: coin_change.py
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Created Time: 2023-07-10
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Author: krahets (krahets@163.com)
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"""
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def coin_change_dp(coins: list[int], amt: int) -> int:
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"""零錢兌換:動態規劃"""
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n = len(coins)
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MAX = amt + 1
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# 初始化 dp 表
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dp = [[0] * (amt + 1) for _ in range(n + 1)]
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# 狀態轉移:首行首列
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for a in range(1, amt + 1):
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dp[0][a] = MAX
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# 狀態轉移:其餘行和列
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for i in range(1, n + 1):
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for a in range(1, amt + 1):
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if coins[i - 1] > a:
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# 若超過目標金額,則不選硬幣 i
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dp[i][a] = dp[i - 1][a]
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else:
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# 不選和選硬幣 i 這兩種方案的較小值
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dp[i][a] = min(dp[i - 1][a], dp[i][a - coins[i - 1]] + 1)
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return dp[n][amt] if dp[n][amt] != MAX else -1
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def coin_change_dp_comp(coins: list[int], amt: int) -> int:
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"""零錢兌換:空間最佳化後的動態規劃"""
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n = len(coins)
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MAX = amt + 1
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# 初始化 dp 表
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dp = [MAX] * (amt + 1)
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dp[0] = 0
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# 狀態轉移
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for i in range(1, n + 1):
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# 正序走訪
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for a in range(1, amt + 1):
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if coins[i - 1] > a:
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# 若超過目標金額,則不選硬幣 i
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dp[a] = dp[a]
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else:
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# 不選和選硬幣 i 這兩種方案的較小值
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dp[a] = min(dp[a], dp[a - coins[i - 1]] + 1)
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return dp[amt] if dp[amt] != MAX else -1
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"""Driver Code"""
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if __name__ == "__main__":
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coins = [1, 2, 5]
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amt = 4
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# 動態規劃
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res = coin_change_dp(coins, amt)
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print(f"湊到目標金額所需的最少硬幣數量為 {res}")
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# 空間最佳化後的動態規劃
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res = coin_change_dp_comp(coins, amt)
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print(f"湊到目標金額所需的最少硬幣數量為 {res}")
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@ -0,0 +1,58 @@
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"""
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File: coin_change_ii.py
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Created Time: 2023-07-10
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Author: krahets (krahets@163.com)
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"""
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def coin_change_ii_dp(coins: list[int], amt: int) -> int:
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"""零錢兌換 II:動態規劃"""
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n = len(coins)
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# 初始化 dp 表
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dp = [[0] * (amt + 1) for _ in range(n + 1)]
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# 初始化首列
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for i in range(n + 1):
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dp[i][0] = 1
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# 狀態轉移
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for i in range(1, n + 1):
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for a in range(1, amt + 1):
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if coins[i - 1] > a:
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# 若超過目標金額,則不選硬幣 i
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dp[i][a] = dp[i - 1][a]
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else:
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# 不選和選硬幣 i 這兩種方案之和
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dp[i][a] = dp[i - 1][a] + dp[i][a - coins[i - 1]]
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return dp[n][amt]
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def coin_change_ii_dp_comp(coins: list[int], amt: int) -> int:
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"""零錢兌換 II:空間最佳化後的動態規劃"""
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n = len(coins)
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# 初始化 dp 表
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dp = [0] * (amt + 1)
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dp[0] = 1
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# 狀態轉移
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for i in range(1, n + 1):
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# 正序走訪
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for a in range(1, amt + 1):
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if coins[i - 1] > a:
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# 若超過目標金額,則不選硬幣 i
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dp[a] = dp[a]
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else:
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# 不選和選硬幣 i 這兩種方案之和
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dp[a] = dp[a] + dp[a - coins[i - 1]]
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return dp[amt]
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"""Driver Code"""
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if __name__ == "__main__":
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coins = [1, 2, 5]
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amt = 5
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# 動態規劃
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res = coin_change_ii_dp(coins, amt)
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print(f"湊出目標金額的硬幣組合數量為 {res}")
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# 空間最佳化後的動態規劃
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res = coin_change_ii_dp_comp(coins, amt)
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print(f"湊出目標金額的硬幣組合數量為 {res}")
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@ -0,0 +1,123 @@
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"""
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File: edit_distancde.py
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Created Time: 2023-07-04
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Author: krahets (krahets@163.com)
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"""
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def edit_distance_dfs(s: str, t: str, i: int, j: int) -> int:
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"""編輯距離:暴力搜尋"""
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# 若 s 和 t 都為空,則返回 0
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if i == 0 and j == 0:
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return 0
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# 若 s 為空,則返回 t 長度
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if i == 0:
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return j
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# 若 t 為空,則返回 s 長度
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if j == 0:
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return i
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# 若兩字元相等,則直接跳過此兩字元
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if s[i - 1] == t[j - 1]:
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return edit_distance_dfs(s, t, i - 1, j - 1)
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# 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 1
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insert = edit_distance_dfs(s, t, i, j - 1)
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delete = edit_distance_dfs(s, t, i - 1, j)
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replace = edit_distance_dfs(s, t, i - 1, j - 1)
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# 返回最少編輯步數
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return min(insert, delete, replace) + 1
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def edit_distance_dfs_mem(s: str, t: str, mem: list[list[int]], i: int, j: int) -> int:
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"""編輯距離:記憶化搜尋"""
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# 若 s 和 t 都為空,則返回 0
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if i == 0 and j == 0:
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return 0
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# 若 s 為空,則返回 t 長度
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if i == 0:
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return j
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# 若 t 為空,則返回 s 長度
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if j == 0:
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return i
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# 若已有記錄,則直接返回之
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if mem[i][j] != -1:
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return mem[i][j]
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# 若兩字元相等,則直接跳過此兩字元
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if s[i - 1] == t[j - 1]:
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return edit_distance_dfs_mem(s, t, mem, i - 1, j - 1)
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# 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 1
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insert = edit_distance_dfs_mem(s, t, mem, i, j - 1)
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delete = edit_distance_dfs_mem(s, t, mem, i - 1, j)
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replace = edit_distance_dfs_mem(s, t, mem, i - 1, j - 1)
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# 記錄並返回最少編輯步數
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mem[i][j] = min(insert, delete, replace) + 1
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return mem[i][j]
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def edit_distance_dp(s: str, t: str) -> int:
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"""編輯距離:動態規劃"""
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n, m = len(s), len(t)
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dp = [[0] * (m + 1) for _ in range(n + 1)]
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# 狀態轉移:首行首列
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for i in range(1, n + 1):
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dp[i][0] = i
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for j in range(1, m + 1):
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dp[0][j] = j
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# 狀態轉移:其餘行和列
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||||
for i in range(1, n + 1):
|
||||
for j in range(1, m + 1):
|
||||
if s[i - 1] == t[j - 1]:
|
||||
# 若兩字元相等,則直接跳過此兩字元
|
||||
dp[i][j] = dp[i - 1][j - 1]
|
||||
else:
|
||||
# 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 1
|
||||
dp[i][j] = min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1
|
||||
return dp[n][m]
|
||||
|
||||
|
||||
def edit_distance_dp_comp(s: str, t: str) -> int:
|
||||
"""編輯距離:空間最佳化後的動態規劃"""
|
||||
n, m = len(s), len(t)
|
||||
dp = [0] * (m + 1)
|
||||
# 狀態轉移:首行
|
||||
for j in range(1, m + 1):
|
||||
dp[j] = j
|
||||
# 狀態轉移:其餘行
|
||||
for i in range(1, n + 1):
|
||||
# 狀態轉移:首列
|
||||
leftup = dp[0] # 暫存 dp[i-1, j-1]
|
||||
dp[0] += 1
|
||||
# 狀態轉移:其餘列
|
||||
for j in range(1, m + 1):
|
||||
temp = dp[j]
|
||||
if s[i - 1] == t[j - 1]:
|
||||
# 若兩字元相等,則直接跳過此兩字元
|
||||
dp[j] = leftup
|
||||
else:
|
||||
# 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 1
|
||||
dp[j] = min(dp[j - 1], dp[j], leftup) + 1
|
||||
leftup = temp # 更新為下一輪的 dp[i-1, j-1]
|
||||
return dp[m]
|
||||
|
||||
|
||||
"""Driver Code"""
|
||||
if __name__ == "__main__":
|
||||
s = "bag"
|
||||
t = "pack"
|
||||
n, m = len(s), len(t)
|
||||
|
||||
# 暴力搜尋
|
||||
res = edit_distance_dfs(s, t, n, m)
|
||||
print(f"將 {s} 更改為 {t} 最少需要編輯 {res} 步")
|
||||
|
||||
# 記憶化搜尋
|
||||
mem = [[-1] * (m + 1) for _ in range(n + 1)]
|
||||
res = edit_distance_dfs_mem(s, t, mem, n, m)
|
||||
print(f"將 {s} 更改為 {t} 最少需要編輯 {res} 步")
|
||||
|
||||
# 動態規劃
|
||||
res = edit_distance_dp(s, t)
|
||||
print(f"將 {s} 更改為 {t} 最少需要編輯 {res} 步")
|
||||
|
||||
# 空間最佳化後的動態規劃
|
||||
res = edit_distance_dp_comp(s, t)
|
||||
print(f"將 {s} 更改為 {t} 最少需要編輯 {res} 步")
|
||||
101
zh-hant/codes/python/chapter_dynamic_programming/knapsack.py
Normal file
101
zh-hant/codes/python/chapter_dynamic_programming/knapsack.py
Normal file
@ -0,0 +1,101 @@
|
||||
"""
|
||||
File: knapsack.py
|
||||
Created Time: 2023-07-03
|
||||
Author: krahets (krahets@163.com)
|
||||
"""
|
||||
|
||||
|
||||
def knapsack_dfs(wgt: list[int], val: list[int], i: int, c: int) -> int:
|
||||
"""0-1 背包:暴力搜尋"""
|
||||
# 若已選完所有物品或背包無剩餘容量,則返回價值 0
|
||||
if i == 0 or c == 0:
|
||||
return 0
|
||||
# 若超過背包容量,則只能選擇不放入背包
|
||||
if wgt[i - 1] > c:
|
||||
return knapsack_dfs(wgt, val, i - 1, c)
|
||||
# 計算不放入和放入物品 i 的最大價值
|
||||
no = knapsack_dfs(wgt, val, i - 1, c)
|
||||
yes = knapsack_dfs(wgt, val, i - 1, c - wgt[i - 1]) + val[i - 1]
|
||||
# 返回兩種方案中價值更大的那一個
|
||||
return max(no, yes)
|
||||
|
||||
|
||||
def knapsack_dfs_mem(
|
||||
wgt: list[int], val: list[int], mem: list[list[int]], i: int, c: int
|
||||
) -> int:
|
||||
"""0-1 背包:記憶化搜尋"""
|
||||
# 若已選完所有物品或背包無剩餘容量,則返回價值 0
|
||||
if i == 0 or c == 0:
|
||||
return 0
|
||||
# 若已有記錄,則直接返回
|
||||
if mem[i][c] != -1:
|
||||
return mem[i][c]
|
||||
# 若超過背包容量,則只能選擇不放入背包
|
||||
if wgt[i - 1] > c:
|
||||
return knapsack_dfs_mem(wgt, val, mem, i - 1, c)
|
||||
# 計算不放入和放入物品 i 的最大價值
|
||||
no = knapsack_dfs_mem(wgt, val, mem, i - 1, c)
|
||||
yes = knapsack_dfs_mem(wgt, val, mem, i - 1, c - wgt[i - 1]) + val[i - 1]
|
||||
# 記錄並返回兩種方案中價值更大的那一個
|
||||
mem[i][c] = max(no, yes)
|
||||
return mem[i][c]
|
||||
|
||||
|
||||
def knapsack_dp(wgt: list[int], val: list[int], cap: int) -> int:
|
||||
"""0-1 背包:動態規劃"""
|
||||
n = len(wgt)
|
||||
# 初始化 dp 表
|
||||
dp = [[0] * (cap + 1) for _ in range(n + 1)]
|
||||
# 狀態轉移
|
||||
for i in range(1, n + 1):
|
||||
for c in range(1, cap + 1):
|
||||
if wgt[i - 1] > c:
|
||||
# 若超過背包容量,則不選物品 i
|
||||
dp[i][c] = dp[i - 1][c]
|
||||
else:
|
||||
# 不選和選物品 i 這兩種方案的較大值
|
||||
dp[i][c] = max(dp[i - 1][c], dp[i - 1][c - wgt[i - 1]] + val[i - 1])
|
||||
return dp[n][cap]
|
||||
|
||||
|
||||
def knapsack_dp_comp(wgt: list[int], val: list[int], cap: int) -> int:
|
||||
"""0-1 背包:空間最佳化後的動態規劃"""
|
||||
n = len(wgt)
|
||||
# 初始化 dp 表
|
||||
dp = [0] * (cap + 1)
|
||||
# 狀態轉移
|
||||
for i in range(1, n + 1):
|
||||
# 倒序走訪
|
||||
for c in range(cap, 0, -1):
|
||||
if wgt[i - 1] > c:
|
||||
# 若超過背包容量,則不選物品 i
|
||||
dp[c] = dp[c]
|
||||
else:
|
||||
# 不選和選物品 i 這兩種方案的較大值
|
||||
dp[c] = max(dp[c], dp[c - wgt[i - 1]] + val[i - 1])
|
||||
return dp[cap]
|
||||
|
||||
|
||||
"""Driver Code"""
|
||||
if __name__ == "__main__":
|
||||
wgt = [10, 20, 30, 40, 50]
|
||||
val = [50, 120, 150, 210, 240]
|
||||
cap = 50
|
||||
n = len(wgt)
|
||||
|
||||
# 暴力搜尋
|
||||
res = knapsack_dfs(wgt, val, n, cap)
|
||||
print(f"不超過背包容量的最大物品價值為 {res}")
|
||||
|
||||
# 記憶化搜尋
|
||||
mem = [[-1] * (cap + 1) for _ in range(n + 1)]
|
||||
res = knapsack_dfs_mem(wgt, val, mem, n, cap)
|
||||
print(f"不超過背包容量的最大物品價值為 {res}")
|
||||
|
||||
# 動態規劃
|
||||
res = knapsack_dp(wgt, val, cap)
|
||||
print(f"不超過背包容量的最大物品價值為 {res}")
|
||||
|
||||
# 空間最佳化後的動態規劃
|
||||
res = knapsack_dp_comp(wgt, val, cap)
|
||||
print(f"不超過背包容量的最大物品價值為 {res}")
|
||||
@ -0,0 +1,43 @@
|
||||
"""
|
||||
File: min_cost_climbing_stairs_dp.py
|
||||
Created Time: 2023-06-30
|
||||
Author: krahets (krahets@163.com)
|
||||
"""
|
||||
|
||||
|
||||
def min_cost_climbing_stairs_dp(cost: list[int]) -> int:
|
||||
"""爬樓梯最小代價:動態規劃"""
|
||||
n = len(cost) - 1
|
||||
if n == 1 or n == 2:
|
||||
return cost[n]
|
||||
# 初始化 dp 表,用於儲存子問題的解
|
||||
dp = [0] * (n + 1)
|
||||
# 初始狀態:預設最小子問題的解
|
||||
dp[1], dp[2] = cost[1], cost[2]
|
||||
# 狀態轉移:從較小子問題逐步求解較大子問題
|
||||
for i in range(3, n + 1):
|
||||
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
|
||||
return dp[n]
|
||||
|
||||
|
||||
def min_cost_climbing_stairs_dp_comp(cost: list[int]) -> int:
|
||||
"""爬樓梯最小代價:空間最佳化後的動態規劃"""
|
||||
n = len(cost) - 1
|
||||
if n == 1 or n == 2:
|
||||
return cost[n]
|
||||
a, b = cost[1], cost[2]
|
||||
for i in range(3, n + 1):
|
||||
a, b = b, min(a, b) + cost[i]
|
||||
return b
|
||||
|
||||
|
||||
"""Driver Code"""
|
||||
if __name__ == "__main__":
|
||||
cost = [0, 1, 10, 1, 1, 1, 10, 1, 1, 10, 1]
|
||||
print(f"輸入樓梯的代價串列為 {cost}")
|
||||
|
||||
res = min_cost_climbing_stairs_dp(cost)
|
||||
print(f"爬完樓梯的最低代價為 {res}")
|
||||
|
||||
res = min_cost_climbing_stairs_dp_comp(cost)
|
||||
print(f"爬完樓梯的最低代價為 {res}")
|
||||
104
zh-hant/codes/python/chapter_dynamic_programming/min_path_sum.py
Normal file
104
zh-hant/codes/python/chapter_dynamic_programming/min_path_sum.py
Normal file
@ -0,0 +1,104 @@
|
||||
"""
|
||||
File: min_path_sum.py
|
||||
Created Time: 2023-07-04
|
||||
Author: krahets (krahets@163.com)
|
||||
"""
|
||||
|
||||
from math import inf
|
||||
|
||||
|
||||
def min_path_sum_dfs(grid: list[list[int]], i: int, j: int) -> int:
|
||||
"""最小路徑和:暴力搜尋"""
|
||||
# 若為左上角單元格,則終止搜尋
|
||||
if i == 0 and j == 0:
|
||||
return grid[0][0]
|
||||
# 若行列索引越界,則返回 +∞ 代價
|
||||
if i < 0 or j < 0:
|
||||
return inf
|
||||
# 計算從左上角到 (i-1, j) 和 (i, j-1) 的最小路徑代價
|
||||
up = min_path_sum_dfs(grid, i - 1, j)
|
||||
left = min_path_sum_dfs(grid, i, j - 1)
|
||||
# 返回從左上角到 (i, j) 的最小路徑代價
|
||||
return min(left, up) + grid[i][j]
|
||||
|
||||
|
||||
def min_path_sum_dfs_mem(
|
||||
grid: list[list[int]], mem: list[list[int]], i: int, j: int
|
||||
) -> int:
|
||||
"""最小路徑和:記憶化搜尋"""
|
||||
# 若為左上角單元格,則終止搜尋
|
||||
if i == 0 and j == 0:
|
||||
return grid[0][0]
|
||||
# 若行列索引越界,則返回 +∞ 代價
|
||||
if i < 0 or j < 0:
|
||||
return inf
|
||||
# 若已有記錄,則直接返回
|
||||
if mem[i][j] != -1:
|
||||
return mem[i][j]
|
||||
# 左邊和上邊單元格的最小路徑代價
|
||||
up = min_path_sum_dfs_mem(grid, mem, i - 1, j)
|
||||
left = min_path_sum_dfs_mem(grid, mem, i, j - 1)
|
||||
# 記錄並返回左上角到 (i, j) 的最小路徑代價
|
||||
mem[i][j] = min(left, up) + grid[i][j]
|
||||
return mem[i][j]
|
||||
|
||||
|
||||
def min_path_sum_dp(grid: list[list[int]]) -> int:
|
||||
"""最小路徑和:動態規劃"""
|
||||
n, m = len(grid), len(grid[0])
|
||||
# 初始化 dp 表
|
||||
dp = [[0] * m for _ in range(n)]
|
||||
dp[0][0] = grid[0][0]
|
||||
# 狀態轉移:首行
|
||||
for j in range(1, m):
|
||||
dp[0][j] = dp[0][j - 1] + grid[0][j]
|
||||
# 狀態轉移:首列
|
||||
for i in range(1, n):
|
||||
dp[i][0] = dp[i - 1][0] + grid[i][0]
|
||||
# 狀態轉移:其餘行和列
|
||||
for i in range(1, n):
|
||||
for j in range(1, m):
|
||||
dp[i][j] = min(dp[i][j - 1], dp[i - 1][j]) + grid[i][j]
|
||||
return dp[n - 1][m - 1]
|
||||
|
||||
|
||||
def min_path_sum_dp_comp(grid: list[list[int]]) -> int:
|
||||
"""最小路徑和:空間最佳化後的動態規劃"""
|
||||
n, m = len(grid), len(grid[0])
|
||||
# 初始化 dp 表
|
||||
dp = [0] * m
|
||||
# 狀態轉移:首行
|
||||
dp[0] = grid[0][0]
|
||||
for j in range(1, m):
|
||||
dp[j] = dp[j - 1] + grid[0][j]
|
||||
# 狀態轉移:其餘行
|
||||
for i in range(1, n):
|
||||
# 狀態轉移:首列
|
||||
dp[0] = dp[0] + grid[i][0]
|
||||
# 狀態轉移:其餘列
|
||||
for j in range(1, m):
|
||||
dp[j] = min(dp[j - 1], dp[j]) + grid[i][j]
|
||||
return dp[m - 1]
|
||||
|
||||
|
||||
"""Driver Code"""
|
||||
if __name__ == "__main__":
|
||||
grid = [[1, 3, 1, 5], [2, 2, 4, 2], [5, 3, 2, 1], [4, 3, 5, 2]]
|
||||
n, m = len(grid), len(grid[0])
|
||||
|
||||
# 暴力搜尋
|
||||
res = min_path_sum_dfs(grid, n - 1, m - 1)
|
||||
print(f"從左上角到右下角的做小路徑和為 {res}")
|
||||
|
||||
# 記憶化搜尋
|
||||
mem = [[-1] * m for _ in range(n)]
|
||||
res = min_path_sum_dfs_mem(grid, mem, n - 1, m - 1)
|
||||
print(f"從左上角到右下角的做小路徑和為 {res}")
|
||||
|
||||
# 動態規劃
|
||||
res = min_path_sum_dp(grid)
|
||||
print(f"從左上角到右下角的做小路徑和為 {res}")
|
||||
|
||||
# 空間最佳化後的動態規劃
|
||||
res = min_path_sum_dp_comp(grid)
|
||||
print(f"從左上角到右下角的做小路徑和為 {res}")
|
||||
@ -0,0 +1,55 @@
|
||||
"""
|
||||
File: unbounded_knapsack.py
|
||||
Created Time: 2023-07-10
|
||||
Author: krahets (krahets@163.com)
|
||||
"""
|
||||
|
||||
|
||||
def unbounded_knapsack_dp(wgt: list[int], val: list[int], cap: int) -> int:
|
||||
"""完全背包:動態規劃"""
|
||||
n = len(wgt)
|
||||
# 初始化 dp 表
|
||||
dp = [[0] * (cap + 1) for _ in range(n + 1)]
|
||||
# 狀態轉移
|
||||
for i in range(1, n + 1):
|
||||
for c in range(1, cap + 1):
|
||||
if wgt[i - 1] > c:
|
||||
# 若超過背包容量,則不選物品 i
|
||||
dp[i][c] = dp[i - 1][c]
|
||||
else:
|
||||
# 不選和選物品 i 這兩種方案的較大值
|
||||
dp[i][c] = max(dp[i - 1][c], dp[i][c - wgt[i - 1]] + val[i - 1])
|
||||
return dp[n][cap]
|
||||
|
||||
|
||||
def unbounded_knapsack_dp_comp(wgt: list[int], val: list[int], cap: int) -> int:
|
||||
"""完全背包:空間最佳化後的動態規劃"""
|
||||
n = len(wgt)
|
||||
# 初始化 dp 表
|
||||
dp = [0] * (cap + 1)
|
||||
# 狀態轉移
|
||||
for i in range(1, n + 1):
|
||||
# 正序走訪
|
||||
for c in range(1, cap + 1):
|
||||
if wgt[i - 1] > c:
|
||||
# 若超過背包容量,則不選物品 i
|
||||
dp[c] = dp[c]
|
||||
else:
|
||||
# 不選和選物品 i 這兩種方案的較大值
|
||||
dp[c] = max(dp[c], dp[c - wgt[i - 1]] + val[i - 1])
|
||||
return dp[cap]
|
||||
|
||||
|
||||
"""Driver Code"""
|
||||
if __name__ == "__main__":
|
||||
wgt = [1, 2, 3]
|
||||
val = [5, 11, 15]
|
||||
cap = 4
|
||||
|
||||
# 動態規劃
|
||||
res = unbounded_knapsack_dp(wgt, val, cap)
|
||||
print(f"不超過背包容量的最大物品價值為 {res}")
|
||||
|
||||
# 空間最佳化後的動態規劃
|
||||
res = unbounded_knapsack_dp_comp(wgt, val, cap)
|
||||
print(f"不超過背包容量的最大物品價值為 {res}")
|
||||
Reference in New Issue
Block a user