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Add Java and C++ code for the chapter of

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divide and conquer.
pull/628/head
krahets 1 year ago
parent fc7bcb615d
commit 1f784dadb0
10 changed files with 331 additions and 6 deletions
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3
codes/cpp/CMakeLists.txt
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@ -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)

3
codes/cpp/chapter_divide_and_conquer/CMakeLists.txt
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@ -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)

46
codes/cpp/chapter_divide_and_conquer/binary_search_recur.cpp
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@ -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<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 */
int binarySearch(vector<int> &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<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);
cout << "目标元素 6 的索引 = " << index << endl;
return 0;
}

51
codes/cpp/chapter_divide_and_conquer/build_tree.cpp
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@ -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<int> &preorder, vector<int> &inorder, unordered_map<int, int> &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<int> &preorder, vector<int> &inorder) {
// 初始化哈希表，存储 inorder 元素到索引的映射
unordered_map<int, int> 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<int> preorder = {3, 9, 2, 1, 7};
vector<int> 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;
}

66
codes/cpp/chapter_divide_and_conquer/hanota.cpp
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@ -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<int> &src, vector<int> &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<int> &src, vector<int> &buf, vector<int> &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<int> &A, vector<int> &B, vector<int> &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<int> A = {5, 4, 3, 2, 1};
vector<int> B = {};
vector<int> 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;
}

45
codes/java/chapter_divide_and_conquer/binary_search_recur.java
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@ -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);
}
}

51
codes/java/chapter_divide_and_conquer/build_tree.java
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@ -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<Integer, Integer> 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<Integer, Integer> 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);
}
}

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codes/java/chapter_divide_and_conquer/hanota.java
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@ -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<Integer> src, List<Integer> 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<Integer> src, List<Integer> buf, List<Integer> 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<Integer> A, List<Integer> B, List<Integer> 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<Integer> A = new ArrayList<>(Arrays.asList(5, 4, 3, 2, 1));
List<Integer> B = new ArrayList<>();
List<Integer> 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);
}
}

11
codes/python/chapter_divide_and_conquer/binary_search_recur.py
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@ -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)

2
docs/chapter_divide_and_conquer/divide_and_conquer.md
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@ -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。

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