From 1f784dadb076862ed66fed36539fa489b5c7b41e Mon Sep 17 00:00:00 2001 From: krahets Date: Mon, 17 Jul 2023 04:20:12 +0800 Subject: [PATCH] Add Java and C++ code for the chapter of divide and conquer. --- codes/cpp/CMakeLists.txt | 3 +- .../chapter_divide_and_conquer/CMakeLists.txt | 3 + .../binary_search_recur.cpp | 46 +++++++++++++ .../chapter_divide_and_conquer/build_tree.cpp | 51 ++++++++++++++ .../cpp/chapter_divide_and_conquer/hanota.cpp | 66 +++++++++++++++++++ .../binary_search_recur.java | 45 +++++++++++++ .../build_tree.java | 51 ++++++++++++++ .../chapter_divide_and_conquer/hanota.java | 59 +++++++++++++++++ .../binary_search_recur.py | 11 ++-- .../divide_and_conquer.md | 2 + 10 files changed, 331 insertions(+), 6 deletions(-) create mode 100644 codes/cpp/chapter_divide_and_conquer/CMakeLists.txt create mode 100644 codes/cpp/chapter_divide_and_conquer/binary_search_recur.cpp create mode 100644 codes/cpp/chapter_divide_and_conquer/build_tree.cpp create mode 100644 codes/cpp/chapter_divide_and_conquer/hanota.cpp create mode 100644 codes/java/chapter_divide_and_conquer/binary_search_recur.java create mode 100644 codes/java/chapter_divide_and_conquer/build_tree.java create mode 100644 codes/java/chapter_divide_and_conquer/hanota.java diff --git a/codes/cpp/CMakeLists.txt b/codes/cpp/CMakeLists.txt index f3d16ae1a..9e3dffc78 100644 --- a/codes/cpp/CMakeLists.txt +++ b/codes/cpp/CMakeLists.txt @@ -12,7 +12,8 @@ add_subdirectory(chapter_hashing) add_subdirectory(chapter_tree) add_subdirectory(chapter_heap) add_subdirectory(chapter_graph) -add_subdirectory(chapter_sorting) add_subdirectory(chapter_searching) +add_subdirectory(chapter_sorting) +add_subdirectory(chapter_divide_and_conquer) add_subdirectory(chapter_backtracking) add_subdirectory(chapter_dynamic_programming) diff --git a/codes/cpp/chapter_divide_and_conquer/CMakeLists.txt b/codes/cpp/chapter_divide_and_conquer/CMakeLists.txt new file mode 100644 index 000000000..38dfff710 --- /dev/null +++ b/codes/cpp/chapter_divide_and_conquer/CMakeLists.txt @@ -0,0 +1,3 @@ +add_executable(binary_search_recur binary_search_recur.cpp) +add_executable(build_tree build_tree.cpp) +add_executable(hanota hanota.cpp) \ No newline at end of file diff --git a/codes/cpp/chapter_divide_and_conquer/binary_search_recur.cpp b/codes/cpp/chapter_divide_and_conquer/binary_search_recur.cpp new file mode 100644 index 000000000..5bb3b248f --- /dev/null +++ b/codes/cpp/chapter_divide_and_conquer/binary_search_recur.cpp @@ -0,0 +1,46 @@ +/** + * File: binary_search_recur.cpp + * Created Time: 2023-07-17 + * Author: krahets (krahets@163.com) + */ + +#include "../utils/common.hpp" + +/* 二分查找：问题 f(i, j) */ +int dfs(vector &nums, int target, int i, int j) { + // 若区间为空，代表无目标元素，则返回 -1 + if (i > j) { + return -1; + } + // 计算中点索引 m + int m = (i + j) / 2; + if (nums[m] < target) { + // 递归子问题 f(m+1, j) + return dfs(nums, target, m + 1, j); + } else if (nums[m] > target) { + // 递归子问题 f(i, m-1) + return dfs(nums, target, i, m - 1); + } else { + // 找到目标元素，返回其索引 + return m; + } +} + +/* 二分查找 */ +int binarySearch(vector &nums, int target) { + int n = nums.size(); + // 求解问题 f(0, n-1) + return dfs(nums, target, 0, n - 1); +} + +/* Driver Code */ +int main() { + int target = 6; + vector nums = {1, 3, 6, 8, 12, 15, 23, 26, 31, 35}; + + // 二分查找（双闭区间） + int index = binarySearch(nums, target); + cout << "目标元素 6 的索引 = " << index << endl; + + return 0; +} \ No newline at end of file diff --git a/codes/cpp/chapter_divide_and_conquer/build_tree.cpp b/codes/cpp/chapter_divide_and_conquer/build_tree.cpp new file mode 100644 index 000000000..51b8fbe24 --- /dev/null +++ b/codes/cpp/chapter_divide_and_conquer/build_tree.cpp @@ -0,0 +1,51 @@ +/** + * File: build_tree.cpp + * Created Time: 2023-07-17 + * Author: Krahets (krahets@163.com) + */ + +#include "../utils/common.hpp" + +/* 构建二叉树：分治 */ +TreeNode *dfs(vector &preorder, vector &inorder, unordered_map &hmap, int i, int l, int r) { + // 子树区间为空时终止 + if (r - l < 0) + return NULL; + // 初始化根节点 + TreeNode *root = new TreeNode(preorder[i]); + // 查询 m ，从而划分左右子树 + int m = hmap[preorder[i]]; + // 子问题：构建左子树 + root->left = dfs(preorder, inorder, hmap, i + 1, l, m - 1); + // 子问题：构建右子树 + root->right = dfs(preorder, inorder, hmap, i + 1 + m - l, m + 1, r); + // 返回根节点 + return root; +} + +/* 构建二叉树 */ +TreeNode *buildTree(vector &preorder, vector &inorder) { + // 初始化哈希表，存储 inorder 元素到索引的映射 + unordered_map hmap; + for (int i = 0; i < inorder.size(); i++) { + hmap[inorder[i]] = i; + } + TreeNode *root = dfs(preorder, inorder, hmap, 0, 0, inorder.size() - 1); + return root; +} + +/* Driver Code */ +int main() { + vector preorder = {3, 9, 2, 1, 7}; + vector inorder = {9, 3, 1, 2, 7}; + cout << "前序遍历 = "; + printVector(preorder); + cout << "中序遍历 = "; + printVector(inorder); + + TreeNode *root = buildTree(preorder, inorder); + cout << "构建的二叉树为：\n"; + printTree(root); + + return 0; +} diff --git a/codes/cpp/chapter_divide_and_conquer/hanota.cpp b/codes/cpp/chapter_divide_and_conquer/hanota.cpp new file mode 100644 index 000000000..308982a0a --- /dev/null +++ b/codes/cpp/chapter_divide_and_conquer/hanota.cpp @@ -0,0 +1,66 @@ +/** + * File: hanota.cpp + * Created Time: 2023-07-17 + * Author: Krahets (krahets@163.com) + */ + +#include "../utils/common.hpp" + +/* 移动一个圆盘 */ +void move(vector &src, vector &tar) { + // 从 src 顶部拿出一个圆盘 + int pan = src.back(); + src.pop_back(); + // 将圆盘放入 tar 顶部 + tar.push_back(pan); +} + +/* 求解汉诺塔：问题 f(i) */ +void dfs(int i, vector &src, vector &buf, vector &tar) { + // 若 src 只剩下一个圆盘，则直接将其移到 tar + if (i == 1) { + move(src, tar); + return; + } + // 子问题 f(i-1) ：将 src 顶部 i-1 个圆盘借助 tar 移到 buf + dfs(i - 1, src, tar, buf); + // 子问题 f(1) ：将 src 剩余一个圆盘移到 tar + move(src, tar); + // 子问题 f(i-1) ：将 buf 顶部 i-1 个圆盘借助 src 移到 tar + dfs(i - 1, buf, src, tar); +} + +/* 求解汉诺塔 */ +void hanota(vector &A, vector &B, vector &C) { + int n = A.size(); + // 将 A 顶部 n 个圆盘借助 B 移到 C + dfs(n, A, B, C); +} + +/* Driver Code */ +int main() { + // 列表尾部是柱子顶部 + vector A = {5, 4, 3, 2, 1}; + vector B = {}; + vector C = {}; + + cout << "初始状态下：\n"; + cout << "A ="; + printVector(A); + cout << "B ="; + printVector(B); + cout << "C ="; + printVector(C); + + hanota(A, B, C); + + cout << "圆盘移动完成后：\n"; + cout << "A ="; + printVector(A); + cout << "B ="; + printVector(B); + cout << "C ="; + printVector(C); + + return 0; +} diff --git a/codes/java/chapter_divide_and_conquer/binary_search_recur.java b/codes/java/chapter_divide_and_conquer/binary_search_recur.java new file mode 100644 index 000000000..1b4f6af8a --- /dev/null +++ b/codes/java/chapter_divide_and_conquer/binary_search_recur.java @@ -0,0 +1,45 @@ +/** + * File: binary_search_recur.java + * Created Time: 2023-07-17 + * Author: krahets (krahets@163.com) + */ + +package chapter_divide_and_conquer; + +public class binary_search_recur { + /* 二分查找：问题 f(i, j) */ + static int dfs(int[] nums, int target, int i, int j) { + // 若区间为空，代表无目标元素，则返回 -1 + if (i > j) { + return -1; + } + // 计算中点索引 m + int m = (i + j) / 2; + if (nums[m] < target) { + // 递归子问题 f(m+1, j) + return dfs(nums, target, m + 1, j); + } else if (nums[m] > target) { + // 递归子问题 f(i, m-1) + return dfs(nums, target, i, m - 1); + } else { + // 找到目标元素，返回其索引 + return m; + } + } + + /* 二分查找 */ + static int binarySearch(int[] nums, int target) { + int n = nums.length; + // 求解问题 f(0, n-1) + return dfs(nums, target, 0, n - 1); + } + + public static void main(String[] args) { + int target = 6; + int[] nums = { 1, 3, 6, 8, 12, 15, 23, 26, 31, 35 }; + + // 二分查找（双闭区间） + int index = binarySearch(nums, target); + System.out.println("目标元素 6 的索引 = " + index); + } +} diff --git a/codes/java/chapter_divide_and_conquer/build_tree.java b/codes/java/chapter_divide_and_conquer/build_tree.java new file mode 100644 index 000000000..082ff5f6c --- /dev/null +++ b/codes/java/chapter_divide_and_conquer/build_tree.java @@ -0,0 +1,51 @@ +/** + * File: build_tree.java + * Created Time: 2023-07-17 + * Author: Krahets (krahets@163.com) + */ + +package chapter_divide_and_conquer; + +import utils.*; +import java.util.*; + +public class build_tree { + /* 构建二叉树：分治 */ + static TreeNode dfs(int[] preorder, int[] inorder, Map hmap, int i, int l, int r) { + // 子树区间为空时终止 + if (r - l < 0) + return null; + // 初始化根节点 + TreeNode root = new TreeNode(preorder[i]); + // 查询 m ，从而划分左右子树 + int m = hmap.get(preorder[i]); + // 子问题：构建左子树 + root.left = dfs(preorder, inorder, hmap, i + 1, l, m - 1); + // 子问题：构建右子树 + root.right = dfs(preorder, inorder, hmap, i + 1 + m - l, m + 1, r); + // 返回根节点 + return root; + } + + /* 构建二叉树 */ + static TreeNode buildTree(int[] preorder, int[] inorder) { + // 初始化哈希表，存储 inorder 元素到索引的映射 + Map hmap = new HashMap<>(); + for (int i = 0; i < inorder.length; i++) { + hmap.put(inorder[i], i); + } + TreeNode root = dfs(preorder, inorder, hmap, 0, 0, inorder.length - 1); + return root; + } + + public static void main(String[] args) { + int[] preorder = { 3, 9, 2, 1, 7 }; + int[] inorder = { 9, 3, 1, 2, 7 }; + System.out.println("前序遍历 = " + Arrays.toString(preorder)); + System.out.println("中序遍历 = " + Arrays.toString(inorder)); + + TreeNode root = buildTree(preorder, inorder); + System.out.println("构建的二叉树为："); + PrintUtil.printTree(root); + } +} diff --git a/codes/java/chapter_divide_and_conquer/hanota.java b/codes/java/chapter_divide_and_conquer/hanota.java new file mode 100644 index 000000000..6f697ef4e --- /dev/null +++ b/codes/java/chapter_divide_and_conquer/hanota.java @@ -0,0 +1,59 @@ +/** + * File: hanota.java + * Created Time: 2023-07-17 + * Author: Krahets (krahets@163.com) + */ + +package chapter_divide_and_conquer; + +import java.util.*; + +public class hanota { + /* 移动一个圆盘 */ + static void move(List src, List tar) { + // 从 src 顶部拿出一个圆盘 + Integer pan = src.remove(src.size() - 1); + // 将圆盘放入 tar 顶部 + tar.add(pan); + } + + /* 求解汉诺塔：问题 f(i) */ + static void dfs(int i, List src, List buf, List tar) { + // 若 src 只剩下一个圆盘，则直接将其移到 tar + if (i == 1) { + move(src, tar); + return; + } + // 子问题 f(i-1) ：将 src 顶部 i-1 个圆盘借助 tar 移到 buf + dfs(i - 1, src, tar, buf); + // 子问题 f(1) ：将 src 剩余一个圆盘移到 tar + move(src, tar); + // 子问题 f(i-1) ：将 buf 顶部 i-1 个圆盘借助 src 移到 tar + dfs(i - 1, buf, src, tar); + } + + /* 求解汉诺塔 */ + static void hanota(List A, List B, List C) { + int n = A.size(); + // 将 A 顶部 n 个圆盘借助 B 移到 C + dfs(n, A, B, C); + } + + public static void main(String[] args) { + // 列表尾部是柱子顶部 + List A = new ArrayList<>(Arrays.asList(5, 4, 3, 2, 1)); + List B = new ArrayList<>(); + List C = new ArrayList<>(); + System.out.println("初始状态下："); + System.out.println("A = " + A); + System.out.println("B = " + B); + System.out.println("C = " + C); + + hanota(A, B, C); + + System.out.println("圆盘移动完成后："); + System.out.println("A = " + A); + System.out.println("B = " + B); + System.out.println("C = " + C); + } +} diff --git a/codes/python/chapter_divide_and_conquer/binary_search_recur.py b/codes/python/chapter_divide_and_conquer/binary_search_recur.py index 26a8884b2..02978201b 100644 --- a/codes/python/chapter_divide_and_conquer/binary_search_recur.py +++ b/codes/python/chapter_divide_and_conquer/binary_search_recur.py @@ -1,22 +1,22 @@ """ File: binary_search_recur.py Created Time: 2023-07-17 -Author: krahets (xisunyy@163.com) +Author: krahets (krahets@163.com) """ def dfs(nums: list[int], target: int, i: int, j: int) -> int: - """二分查找：分治""" - # 若区间为空，代表未找到目标元素，则返回 -1 + """二分查找：问题 f(i, j)""" + # 若区间为空，代表无目标元素，则返回 -1 if i > j: return -1 # 计算中点索引 m m = (i + j) // 2 if nums[m] < target: - # 此情况说明 target 在区间 [m+1, j] 中，递归解决该子问题 + # 递归子问题 f(m+1, j) return dfs(nums, target, m + 1, j) elif nums[m] > target: - # 此情况说明 target 在区间 [i, m-1] 中，递归解决该子问题 + # 递归子问题 f(i, m-1) return dfs(nums, target, i, m - 1) else: # 找到目标元素，返回其索引 @@ -26,6 +26,7 @@ def dfs(nums: list[int], target: int, i: int, j: int) -> int: def binary_search(nums: list[int], target: int) -> int: """二分查找""" n = len(nums) + # 求解问题 f(0, n-1) return dfs(nums, target, 0, n - 1) diff --git a/docs/chapter_divide_and_conquer/divide_and_conquer.md b/docs/chapter_divide_and_conquer/divide_and_conquer.md index c0886b7db..c6e9e8f1d 100644 --- a/docs/chapter_divide_and_conquer/divide_and_conquer.md +++ b/docs/chapter_divide_and_conquer/divide_and_conquer.md @@ -12,6 +12,8 @@ ![归并排序的分治策略](divide_and_conquer.assets/divide_and_conquer_merge_sort.png) +## 如何判断分治问题 + 一个问题是否适合使用分治解决，通常可以参考以下几个判断依据： 1. **问题可以被分解**：原问题可以被分解成规模更小、类似的子问题，以及能够以相同方式递归地进行划分。