diff --git a/codes/ruby/chapter_divide_and_conquer/binary_search_recur.rb b/codes/ruby/chapter_divide_and_conquer/binary_search_recur.rb new file mode 100644 index 000000000..214a7cbf5 --- /dev/null +++ b/codes/ruby/chapter_divide_and_conquer/binary_search_recur.rb @@ -0,0 +1,42 @@ +=begin +File: binary_search_recur.rb +Created Time: 2024-05-13 +Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com) +=end + +### 二分查找:问题 f(i, j) ### +def dfs(nums, target, i, j) + # 若区间为空,代表无目标元素,则返回 -1 + return -1 if i > j + + # 计算中点索引 m + m = (i + j) / 2 + + if nums[m] < target + # 递归子问题 f(m+1, j) + return dfs(nums, target, m + 1, j) + elsif 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 + end +end + +### 二分查找 ### +def binary_search(nums, target) + n = nums.length + # 求解问题 f(0, n-1) + dfs(nums, target, 0, n - 1) +end + +### Driver Code ### +if __FILE__ == $0 + target = 6 + nums = [1, 3, 6, 8, 12, 15, 23, 26, 31, 35] + + # 二分查找(双闭区间) + index = binary_search(nums, target) + puts "目标元素 6 的索引 = #{index}" +end diff --git a/codes/ruby/chapter_divide_and_conquer/build_tree.rb b/codes/ruby/chapter_divide_and_conquer/build_tree.rb new file mode 100644 index 000000000..3d562c820 --- /dev/null +++ b/codes/ruby/chapter_divide_and_conquer/build_tree.rb @@ -0,0 +1,46 @@ +=begin +File: build_tree.rb +Created Time: 2024-05-13 +Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com) +=end + +require_relative '../utils/tree_node' +require_relative '../utils/print_util' + +### 构建二叉树:分治 ### +def dfs(preorder, inorder_map, i, l, r) + # 子树区间为空时终止 + return if r - l < 0 + + # 初始化根节点 + root = TreeNode.new(preorder[i]) + # 查询 m ,从而划分左右子树 + m = inorder_map[preorder[i]] + # 子问题:构建左子树 + root.left = dfs(preorder, inorder_map, i + 1, l, m - 1) + # 子问题:构建右子树 + root.right = dfs(preorder, inorder_map, i + 1 + m - l, m + 1, r) + + # 返回根节点 + root +end + +### 构建二叉树 ### +def build_tree(preorder, inorder) + # 初始化哈希表,存储 inorder 元素到索引的映射 + inorder_map = {} + inorder.each_with_index { |val, i| inorder_map[val] = i } + dfs(preorder, inorder_map, 0, 0, inorder.length - 1) +end + +### Driver Code ### +if __FILE__ == $0 + preorder = [3, 9, 2, 1, 7] + inorder = [9, 3, 1, 2, 7] + puts "前序遍历 = #{preorder}" + puts "中序遍历 = #{inorder}" + + root = build_tree(preorder, inorder) + puts "构建的二叉树为:" + print_tree(root) +end diff --git a/codes/ruby/chapter_divide_and_conquer/hanota.rb b/codes/ruby/chapter_divide_and_conquer/hanota.rb new file mode 100644 index 000000000..456ccbd4e --- /dev/null +++ b/codes/ruby/chapter_divide_and_conquer/hanota.rb @@ -0,0 +1,55 @@ +=begin +File: hanota.rb +Created Time: 2024-05-13 +Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com) +=end + +### 移动一个圆盘 ### +def move(src, tar) + # 从 src 顶部拿出一个圆盘 + pan = src.pop + # 将圆盘放入 tar 顶部 + tar << pan +end + +### 求解汉诺塔问题 f(i) ### +def dfs(i, src, buf, 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 + end + + # 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) +end + +### 求解汉诺塔问题 ### +def solve_hanota(_A, _B, _C) + n = _A.length + # 将 A 顶部 n 个圆盘借助 B 移到 C + dfs(n, _A, _B, _C) +end + +### Driver Code ### +if __FILE__ == $0 + # 列表尾部是柱子顶部 + A = [5, 4, 3, 2, 1] + B = [] + C = [] + puts "初始状态下:" + puts "A = #{A}" + puts "B = #{B}" + puts "C = #{C}" + + solve_hanota(A, B, C) + + puts "圆盘移动完成后:" + puts "A = #{A}" + puts "B = #{B}" + puts "C = #{C}" +end diff --git a/codes/rust/chapter_divide_and_conquer/hanota.rs b/codes/rust/chapter_divide_and_conquer/hanota.rs index 705f817b9..92880a0bc 100644 --- a/codes/rust/chapter_divide_and_conquer/hanota.rs +++ b/codes/rust/chapter_divide_and_conquer/hanota.rs @@ -9,7 +9,7 @@ /* 移动一个圆盘 */ fn move_pan(src: &mut Vec, tar: &mut Vec) { // 从 src 顶部拿出一个圆盘 - let pan = src.remove(src.len() - 1); + let pan = src.pop().unwrap(); // 将圆盘放入 tar 顶部 tar.push(pan); } diff --git a/codes/rust/chapter_sorting/bubble_sort.rs b/codes/rust/chapter_sorting/bubble_sort.rs index d7fde10ae..2557a222f 100644 --- a/codes/rust/chapter_sorting/bubble_sort.rs +++ b/codes/rust/chapter_sorting/bubble_sort.rs @@ -14,9 +14,7 @@ fn bubble_sort(nums: &mut [i32]) { for j in 0..i { if nums[j] > nums[j + 1] { // 交换 nums[j] 与 nums[j + 1] - let tmp = nums[j]; - nums[j] = nums[j + 1]; - nums[j + 1] = tmp; + nums.swap(j, j + 1); } } } @@ -31,9 +29,7 @@ fn bubble_sort_with_flag(nums: &mut [i32]) { for j in 0..i { if nums[j] > nums[j + 1] { // 交换 nums[j] 与 nums[j + 1] - let tmp = nums[j]; - nums[j] = nums[j + 1]; - nums[j + 1] = tmp; + nums.swap(j, j + 1); flag = true; // 记录交换元素 } } diff --git a/codes/rust/chapter_sorting/bucket_sort.rs b/codes/rust/chapter_sorting/bucket_sort.rs index 5bec09451..02e09abf2 100644 --- a/codes/rust/chapter_sorting/bucket_sort.rs +++ b/codes/rust/chapter_sorting/bucket_sort.rs @@ -12,7 +12,7 @@ fn bucket_sort(nums: &mut [f64]) { let k = nums.len() / 2; let mut buckets = vec![vec![]; k]; // 1. 将数组元素分配到各个桶中 - for &mut num in &mut *nums { + for &num in nums.iter() { // 输入数据范围为 [0, 1),使用 num * k 映射到索引范围 [0, k-1] let i = (num * k as f64) as usize; // 将 num 添加进桶 i @@ -25,8 +25,8 @@ fn bucket_sort(nums: &mut [f64]) { } // 3. 遍历桶合并结果 let mut i = 0; - for bucket in &mut buckets { - for &mut num in bucket { + for bucket in buckets.iter() { + for &num in bucket.iter() { nums[i] = num; i += 1; } diff --git a/codes/rust/chapter_sorting/counting_sort.rs b/codes/rust/chapter_sorting/counting_sort.rs index ca4ed0de3..c6a5d92cd 100644 --- a/codes/rust/chapter_sorting/counting_sort.rs +++ b/codes/rust/chapter_sorting/counting_sort.rs @@ -10,11 +10,11 @@ include!("../include/include.rs"); // 简单实现,无法用于排序对象 fn counting_sort_naive(nums: &mut [i32]) { // 1. 统计数组最大元素 m - let m = *nums.into_iter().max().unwrap(); + let m = *nums.iter().max().unwrap(); // 2. 统计各数字的出现次数 // counter[num] 代表 num 的出现次数 let mut counter = vec![0; m as usize + 1]; - for &num in &*nums { + for &num in nums.iter() { counter[num as usize] += 1; } // 3. 遍历 counter ,将各元素填入原数组 nums @@ -31,16 +31,16 @@ fn counting_sort_naive(nums: &mut [i32]) { // 完整实现,可排序对象,并且是稳定排序 fn counting_sort(nums: &mut [i32]) { // 1. 统计数组最大元素 m - let m = *nums.into_iter().max().unwrap(); + let m = *nums.iter().max().unwrap() as usize; // 2. 统计各数字的出现次数 // counter[num] 代表 num 的出现次数 - let mut counter = vec![0; m as usize + 1]; - for &num in &*nums { + let mut counter = vec![0; m + 1]; + for &num in nums.iter() { counter[num as usize] += 1; } // 3. 求 counter 的前缀和,将“出现次数”转换为“尾索引” // 即 counter[num]-1 是 num 在 res 中最后一次出现的索引 - for i in 0..m as usize { + for i in 0..m { counter[i + 1] += counter[i]; } // 4. 倒序遍历 nums ,将各元素填入结果数组 res @@ -53,9 +53,7 @@ fn counting_sort(nums: &mut [i32]) { counter[num as usize] -= 1; // 令前缀和自减 1 ,得到下次放置 num 的索引 } // 使用结果数组 res 覆盖原数组 nums - for i in 0..n { - nums[i] = res[i]; - } + nums.copy_from_slice(&res) } /* Driver Code */ diff --git a/codes/rust/chapter_sorting/heap_sort.rs b/codes/rust/chapter_sorting/heap_sort.rs index d01202b4c..55ab52428 100644 --- a/codes/rust/chapter_sorting/heap_sort.rs +++ b/codes/rust/chapter_sorting/heap_sort.rs @@ -24,9 +24,7 @@ fn sift_down(nums: &mut [i32], n: usize, mut i: usize) { break; } // 交换两节点 - let temp = nums[i]; - nums[i] = nums[ma]; - nums[ma] = temp; + nums.swap(i, ma); // 循环向下堆化 i = ma; } @@ -35,15 +33,13 @@ fn sift_down(nums: &mut [i32], n: usize, mut i: usize) { /* 堆排序 */ fn heap_sort(nums: &mut [i32]) { // 建堆操作:堆化除叶节点以外的其他所有节点 - for i in (0..=nums.len() / 2 - 1).rev() { + for i in (0..nums.len() / 2).rev() { sift_down(nums, nums.len(), i); } // 从堆中提取最大元素,循环 n-1 轮 - for i in (1..=nums.len() - 1).rev() { + for i in (1..nums.len()).rev() { // 交换根节点与最右叶节点(交换首元素与尾元素) - let tmp = nums[0]; - nums[0] = nums[i]; - nums[i] = tmp; + nums.swap(0, i); // 以根节点为起点,从顶至底进行堆化 sift_down(nums, i, 0); }