feat: add ruby code - chapter "divide and conquer" (#1361)

codes/ruby/chapter_divide_and_conquer/binary_search_recur.rb

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+=begin
+File: binary_search_recur.rb
+Created Time: 2024-05-13
+Author: Xuan Khoa Tu Nguyen ([email protected])
+=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

codes/ruby/chapter_divide_and_conquer/build_tree.rb

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+=begin
+File: build_tree.rb
+Created Time: 2024-05-13
+Author: Xuan Khoa Tu Nguyen ([email protected])
+=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

codes/ruby/chapter_divide_and_conquer/hanota.rb

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+=begin
+File: hanota.rb
+Created Time: 2024-05-13
+Author: Xuan Khoa Tu Nguyen ([email protected])
+=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