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