mirror of
https://github.com/krahets/hello-algo.git
synced 2024-12-25 01:06:30 +08:00
Add Java and C++ code for the chapter of
divide and conquer.
This commit is contained in:
parent
fc7bcb615d
commit
1f784dadb0
10 changed files with 331 additions and 6 deletions
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@ -12,7 +12,8 @@ add_subdirectory(chapter_hashing)
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add_subdirectory(chapter_tree)
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add_subdirectory(chapter_heap)
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add_subdirectory(chapter_graph)
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add_subdirectory(chapter_sorting)
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add_subdirectory(chapter_searching)
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add_subdirectory(chapter_sorting)
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add_subdirectory(chapter_divide_and_conquer)
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add_subdirectory(chapter_backtracking)
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add_subdirectory(chapter_dynamic_programming)
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3
codes/cpp/chapter_divide_and_conquer/CMakeLists.txt
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3
codes/cpp/chapter_divide_and_conquer/CMakeLists.txt
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@ -0,0 +1,3 @@
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add_executable(binary_search_recur binary_search_recur.cpp)
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add_executable(build_tree build_tree.cpp)
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add_executable(hanota hanota.cpp)
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46
codes/cpp/chapter_divide_and_conquer/binary_search_recur.cpp
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46
codes/cpp/chapter_divide_and_conquer/binary_search_recur.cpp
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/**
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* File: binary_search_recur.cpp
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* Created Time: 2023-07-17
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* Author: krahets (krahets@163.com)
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*/
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#include "../utils/common.hpp"
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/* 二分查找:问题 f(i, j) */
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int dfs(vector<int> &nums, int target, int i, int j) {
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// 若区间为空,代表无目标元素,则返回 -1
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if (i > j) {
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return -1;
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}
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// 计算中点索引 m
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int m = (i + j) / 2;
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if (nums[m] < target) {
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// 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j);
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} else if (nums[m] > target) {
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// 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1);
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} else {
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// 找到目标元素,返回其索引
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return m;
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}
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}
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/* 二分查找 */
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int binarySearch(vector<int> &nums, int target) {
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int n = nums.size();
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// 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 1);
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}
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/* Driver Code */
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int main() {
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int target = 6;
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vector<int> nums = {1, 3, 6, 8, 12, 15, 23, 26, 31, 35};
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// 二分查找(双闭区间)
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int index = binarySearch(nums, target);
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cout << "目标元素 6 的索引 = " << index << endl;
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return 0;
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}
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51
codes/cpp/chapter_divide_and_conquer/build_tree.cpp
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51
codes/cpp/chapter_divide_and_conquer/build_tree.cpp
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@ -0,0 +1,51 @@
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/**
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* File: build_tree.cpp
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* Created Time: 2023-07-17
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* Author: Krahets (krahets@163.com)
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*/
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#include "../utils/common.hpp"
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/* 构建二叉树:分治 */
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TreeNode *dfs(vector<int> &preorder, vector<int> &inorder, unordered_map<int, int> &hmap, int i, int l, int r) {
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// 子树区间为空时终止
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if (r - l < 0)
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return NULL;
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// 初始化根节点
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TreeNode *root = new TreeNode(preorder[i]);
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// 查询 m ,从而划分左右子树
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int m = hmap[preorder[i]];
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// 子问题:构建左子树
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root->left = dfs(preorder, inorder, hmap, i + 1, l, m - 1);
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// 子问题:构建右子树
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root->right = dfs(preorder, inorder, hmap, i + 1 + m - l, m + 1, r);
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// 返回根节点
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return root;
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}
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/* 构建二叉树 */
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TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
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// 初始化哈希表,存储 inorder 元素到索引的映射
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unordered_map<int, int> hmap;
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for (int i = 0; i < inorder.size(); i++) {
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hmap[inorder[i]] = i;
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}
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TreeNode *root = dfs(preorder, inorder, hmap, 0, 0, inorder.size() - 1);
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return root;
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}
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/* Driver Code */
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int main() {
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vector<int> preorder = {3, 9, 2, 1, 7};
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vector<int> inorder = {9, 3, 1, 2, 7};
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cout << "前序遍历 = ";
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printVector(preorder);
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cout << "中序遍历 = ";
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printVector(inorder);
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TreeNode *root = buildTree(preorder, inorder);
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cout << "构建的二叉树为:\n";
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printTree(root);
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return 0;
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}
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66
codes/cpp/chapter_divide_and_conquer/hanota.cpp
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66
codes/cpp/chapter_divide_and_conquer/hanota.cpp
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/**
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* File: hanota.cpp
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* Created Time: 2023-07-17
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* Author: Krahets (krahets@163.com)
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*/
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#include "../utils/common.hpp"
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/* 移动一个圆盘 */
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void move(vector<int> &src, vector<int> &tar) {
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// 从 src 顶部拿出一个圆盘
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int pan = src.back();
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src.pop_back();
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// 将圆盘放入 tar 顶部
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tar.push_back(pan);
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}
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/* 求解汉诺塔:问题 f(i) */
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void dfs(int i, vector<int> &src, vector<int> &buf, vector<int> &tar) {
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// 若 src 只剩下一个圆盘,则直接将其移到 tar
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if (i == 1) {
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move(src, tar);
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return;
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}
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// 子问题 f(i-1) :将 src 顶部 i-1 个圆盘借助 tar 移到 buf
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dfs(i - 1, src, tar, buf);
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// 子问题 f(1) :将 src 剩余一个圆盘移到 tar
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move(src, tar);
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// 子问题 f(i-1) :将 buf 顶部 i-1 个圆盘借助 src 移到 tar
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dfs(i - 1, buf, src, tar);
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}
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/* 求解汉诺塔 */
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void hanota(vector<int> &A, vector<int> &B, vector<int> &C) {
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int n = A.size();
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// 将 A 顶部 n 个圆盘借助 B 移到 C
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dfs(n, A, B, C);
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}
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/* Driver Code */
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int main() {
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// 列表尾部是柱子顶部
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vector<int> A = {5, 4, 3, 2, 1};
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vector<int> B = {};
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vector<int> C = {};
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cout << "初始状态下:\n";
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cout << "A =";
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printVector(A);
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cout << "B =";
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printVector(B);
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cout << "C =";
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printVector(C);
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hanota(A, B, C);
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cout << "圆盘移动完成后:\n";
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cout << "A =";
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printVector(A);
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cout << "B =";
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printVector(B);
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cout << "C =";
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printVector(C);
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return 0;
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}
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/**
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* File: binary_search_recur.java
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* Created Time: 2023-07-17
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* Author: krahets (krahets@163.com)
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*/
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package chapter_divide_and_conquer;
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public class binary_search_recur {
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/* 二分查找:问题 f(i, j) */
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static int dfs(int[] nums, int target, int i, int j) {
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// 若区间为空,代表无目标元素,则返回 -1
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if (i > j) {
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return -1;
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}
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// 计算中点索引 m
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int m = (i + j) / 2;
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if (nums[m] < target) {
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// 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j);
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} else if (nums[m] > target) {
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// 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1);
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} else {
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// 找到目标元素,返回其索引
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return m;
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}
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}
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/* 二分查找 */
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static int binarySearch(int[] nums, int target) {
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int n = nums.length;
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// 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 1);
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}
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public static void main(String[] args) {
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int target = 6;
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int[] nums = { 1, 3, 6, 8, 12, 15, 23, 26, 31, 35 };
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// 二分查找(双闭区间)
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int index = binarySearch(nums, target);
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System.out.println("目标元素 6 的索引 = " + index);
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}
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}
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51
codes/java/chapter_divide_and_conquer/build_tree.java
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51
codes/java/chapter_divide_and_conquer/build_tree.java
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/**
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* File: build_tree.java
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* Created Time: 2023-07-17
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* Author: Krahets (krahets@163.com)
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*/
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package chapter_divide_and_conquer;
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import utils.*;
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import java.util.*;
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public class build_tree {
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/* 构建二叉树:分治 */
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static TreeNode dfs(int[] preorder, int[] inorder, Map<Integer, Integer> hmap, int i, int l, int r) {
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// 子树区间为空时终止
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if (r - l < 0)
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return null;
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// 初始化根节点
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TreeNode root = new TreeNode(preorder[i]);
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// 查询 m ,从而划分左右子树
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int m = hmap.get(preorder[i]);
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// 子问题:构建左子树
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root.left = dfs(preorder, inorder, hmap, i + 1, l, m - 1);
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// 子问题:构建右子树
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root.right = dfs(preorder, inorder, hmap, i + 1 + m - l, m + 1, r);
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// 返回根节点
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return root;
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}
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/* 构建二叉树 */
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static TreeNode buildTree(int[] preorder, int[] inorder) {
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// 初始化哈希表,存储 inorder 元素到索引的映射
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Map<Integer, Integer> hmap = new HashMap<>();
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for (int i = 0; i < inorder.length; i++) {
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hmap.put(inorder[i], i);
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}
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TreeNode root = dfs(preorder, inorder, hmap, 0, 0, inorder.length - 1);
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return root;
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}
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public static void main(String[] args) {
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int[] preorder = { 3, 9, 2, 1, 7 };
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int[] inorder = { 9, 3, 1, 2, 7 };
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System.out.println("前序遍历 = " + Arrays.toString(preorder));
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System.out.println("中序遍历 = " + Arrays.toString(inorder));
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TreeNode root = buildTree(preorder, inorder);
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System.out.println("构建的二叉树为:");
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PrintUtil.printTree(root);
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}
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}
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59
codes/java/chapter_divide_and_conquer/hanota.java
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59
codes/java/chapter_divide_and_conquer/hanota.java
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/**
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* File: hanota.java
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* Created Time: 2023-07-17
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* Author: Krahets (krahets@163.com)
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*/
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package chapter_divide_and_conquer;
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import java.util.*;
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public class hanota {
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/* 移动一个圆盘 */
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static void move(List<Integer> src, List<Integer> tar) {
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// 从 src 顶部拿出一个圆盘
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Integer pan = src.remove(src.size() - 1);
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// 将圆盘放入 tar 顶部
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tar.add(pan);
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}
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/* 求解汉诺塔:问题 f(i) */
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static void dfs(int i, List<Integer> src, List<Integer> buf, List<Integer> tar) {
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// 若 src 只剩下一个圆盘,则直接将其移到 tar
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if (i == 1) {
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move(src, tar);
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return;
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}
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// 子问题 f(i-1) :将 src 顶部 i-1 个圆盘借助 tar 移到 buf
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dfs(i - 1, src, tar, buf);
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// 子问题 f(1) :将 src 剩余一个圆盘移到 tar
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move(src, tar);
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// 子问题 f(i-1) :将 buf 顶部 i-1 个圆盘借助 src 移到 tar
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dfs(i - 1, buf, src, tar);
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}
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/* 求解汉诺塔 */
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static void hanota(List<Integer> A, List<Integer> B, List<Integer> C) {
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int n = A.size();
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// 将 A 顶部 n 个圆盘借助 B 移到 C
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dfs(n, A, B, C);
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}
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public static void main(String[] args) {
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// 列表尾部是柱子顶部
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List<Integer> A = new ArrayList<>(Arrays.asList(5, 4, 3, 2, 1));
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List<Integer> B = new ArrayList<>();
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List<Integer> C = new ArrayList<>();
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System.out.println("初始状态下:");
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System.out.println("A = " + A);
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System.out.println("B = " + B);
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System.out.println("C = " + C);
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hanota(A, B, C);
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System.out.println("圆盘移动完成后:");
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System.out.println("A = " + A);
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System.out.println("B = " + B);
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System.out.println("C = " + C);
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}
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}
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"""
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File: binary_search_recur.py
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Created Time: 2023-07-17
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Author: krahets (xisunyy@163.com)
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Author: krahets (krahets@163.com)
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"""
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def dfs(nums: list[int], target: int, i: int, j: int) -> int:
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"""二分查找:分治"""
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# 若区间为空,代表未找到目标元素,则返回 -1
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"""二分查找:问题 f(i, j)"""
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# 若区间为空,代表无目标元素,则返回 -1
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if i > j:
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return -1
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# 计算中点索引 m
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m = (i + j) // 2
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if nums[m] < target:
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# 此情况说明 target 在区间 [m+1, j] 中,递归解决该子问题
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# 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j)
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elif nums[m] > target:
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# 此情况说明 target 在区间 [i, m-1] 中,递归解决该子问题
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# 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1)
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else:
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# 找到目标元素,返回其索引
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def binary_search(nums: list[int], target: int) -> int:
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"""二分查找"""
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n = len(nums)
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# 求解问题 f(0, n-1)
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return dfs(nums, target, 0, n - 1)
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![归并排序的分治策略](divide_and_conquer.assets/divide_and_conquer_merge_sort.png)
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## 如何判断分治问题
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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:
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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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