feat: Add Ruby code - chapter "computational complexity" (#1212)

* feat: add ruby code - chapter computational complexity

* feat: add ruby code blocks

codes/ruby/chapter_computational_complexity/iteration.rb

@@ -0,0 +1,78 @@
+=begin
+File: iteration.rb
+Created Time: 2024-03-30
+Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
+=end
+
+### for 循环 ###
+def for_loop(n)
+  res = 0
+
+  # 循环求和 1, 2, ..., n-1, n
+  for i in 1..n
+    res += i
+  end
+
+  res
+end
+
+### while 循环 ###
+def while_loop(n)
+  res = 0
+  i = 1 # 初始化条件变量
+
+  # 循环求和 1, 2, ..., n-1, n
+  while i <= n
+    res += i
+    i += 1 # 更新条件变量
+  end
+
+  res
+end
+
+### while 循环（两次更新）###
+def while_loop_ii(n)
+  res = 0
+  i = 1 # 初始化条件变量
+
+  # 循环求和 1, 4, 10, ...
+  while i <= n
+    res += i
+    # 更新条件变量
+    i += 1
+    i *= 2
+  end
+
+  res
+end
+
+### 双层 for 循环 ###
+def nested_for_loop(n)
+  res = ""
+
+  # 循环 i = 1, 2, ..., n-1, n
+  for i in 1..n
+    # 循环 j = 1, 2, ..., n-1, n
+    for j in 1..n
+      res += "(#{i}, #{j}), "
+    end
+  end
+
+  res
+end
+
+### Driver Code ###
+
+n = 5
+
+res = for_loop n
+puts "\nfor 循环的求和结果 res = #{res}"
+
+res = while_loop n
+puts "\nwhile 循环的求和结果 res = #{res}"
+
+res = while_loop_ii n
+puts "\nwhile 循环（两次更新）求和结果 res = #{res}"
+
+res = nested_for_loop n
+puts "\n双层 for 循环的遍历结果 #{res}"

codes/ruby/chapter_computational_complexity/recursion.rb

@@ -0,0 +1,69 @@
+=begin
+File: recursion.rb
+Created Time: 2024-03-30
+Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
+=end
+
+### 递归 ###
+def recur(n)
+  # 终止条件
+  return 1 if n == 1
+  # 递：递归调用
+  res = recur n - 1
+  # 归：返回结果
+  n + res
+end
+
+### 使用迭代模拟递归 ###
+def for_loop_recur(n)
+  # 使用一个显式的栈来模拟系统调用栈
+  stack = []
+  res = 0
+
+  # 递：递归调用
+  for i in n.downto(0)
+    # 通过“入栈操作”模拟“递”
+    stack << i
+  end
+  # 归：返回结果
+  while !stack.empty?
+    res += stack.pop
+  end
+
+  # res = 1+2+3+...+n
+  res
+end
+
+### 尾递归 ###
+def tail_recur(n, res)
+  # 终止条件
+  return res if n == 0
+  # 尾递归调用
+  tail_recur n - 1, res + n
+end
+
+### 斐波那契数列：递归 ###
+def fib(n)
+  # 终止条件 f(1) = 0, f(2) = 1
+  return n - 1 if n == 1 || n == 2
+  # 递归调用 f(n) = f(n-1) + f(n-2)
+  res = fib(n - 1) + fib(n - 2)
+  # 返回结果 f(n)
+  res
+end
+
+### Driver Code ###
+
+n = 5
+
+res = recur n
+puts "\n递归函数的求和结果 res = #{res}"
+
+res = for_loop_recur n
+puts "\n使用迭代模拟递归求和结果 res = #{res}"
+
+res = tail_recur n, 0
+puts "\n尾递归函数的求和结果 res = #{res}"
+
+res = fib n
+puts "\n斐波那契数列的第 #{n} 项为 #{res}"

codes/ruby/chapter_computational_complexity/space_complexity.rb

@@ -0,0 +1,91 @@
+=begin
+File: space_complexity.rb
+Created Time: 2024-03-30
+Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
+=end
+
+require_relative '../utils/list_node'
+require_relative '../utils/tree_node'
+require_relative '../utils/print_util'
+
+### 函数 ###
+def function
+  # 执行某些操作
+  0
+end
+
+### 常数阶 ###
+def constant(n)
+  # 常量、变量、对象占用 O(1) 空间
+  a = 0
+  nums = [0] * 10000
+  node = ListNode.new
+
+  # 循环中的变量占用 O(1) 空间
+  (0...n).each { c = 0 }
+  # 循环中的函数占用 O(1) 空间
+  (0...n).each { function }
+end
+
+### 线性阶 ###
+def linear(n)
+  # 长度为 n 的列表占用 O(n) 空间
+  nums = Array.new n, 0
+
+  # 长度为 n 的哈希表占用 O(n) 空间
+  hmap = {}
+  for i in 0...n
+    hmap[i] = i.to_s
+  end
+end
+
+### 线性阶（递归实现）###
+def linear_recur(n)
+  puts "递归 n = #{n}"
+  return if n == 1
+  linear_recur n - 1
+end
+
+### 平方阶 ###
+def quadratic(n)
+  # 二维列表占用 O(n^2) 空间
+  Array.new(n) { Array.new n, 0 }
+end
+
+### 平方阶（递归实现）###
+def quadratic_recur(n)
+  return 0 unless n > 0
+
+  # 数组 nums 长度为 n, n-1, ..., 2, 1
+  nums = Array.new n, 0
+  quadratic_recur n - 1
+end
+
+### 指数阶（建立满二叉树）###
+def build_tree(n)
+  return if n == 0
+
+  TreeNode.new.tap do |root|
+    root.left = build_tree n - 1
+    root.right = build_tree n - 1
+  end
+end
+
+### Driver Code ###
+
+n = 5
+
+# 常数阶
+constant n
+
+# 线性阶
+linear n
+linear_recur n
+
+# 平方阶
+quadratic n
+quadratic_recur n
+
+# 指数阶
+root = build_tree n
+print_tree root

codes/ruby/chapter_computational_complexity/time_complexity.rb

@@ -0,0 +1,164 @@
+=begin
+File: time_complexity.rb
+Created Time: 2024-03-30
+Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
+=end
+
+### 常数阶 ###
+def constant(n)
+  count = 0
+  size = 100000
+
+  (0...size).each { count += 1 }
+
+  count
+end
+
+### 线性阶 ###
+def linear(n)
+  count = 0
+  (0...n).each { count += 1 }
+  count
+end
+
+### 线性阶（遍历数组）###
+def array_traversal(nums)
+  count = 0
+
+  # 循环次数与数组长度成正比
+  for num in nums
+    count += 1
+  end
+
+  count
+end
+
+### 平方阶 ###
+def quadratic(n)
+  count = 0
+
+  # 循环次数与数据大小 n 成平方关系
+  for i in 0...n
+    for j in 0...n
+      count += 1
+    end
+  end
+
+  count
+end
+
+### 平方阶（冒泡排序）###
+def bubble_sort(nums)
+  count = 0  # 计数器
+
+  # 外循环：未排序区间为 [0, i]
+  for i in (nums.length - 1).downto(0)
+    # 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
+    for j in 0...i
+      if nums[j] > nums[j + 1]
+        # 交换 nums[j] 与 nums[j + 1]
+        tmp = nums[j]
+        nums[j] = nums[j + 1]
+        nums[j + 1] = tmp
+        count += 3 # 元素交换包含 3 个单元操作
+      end
+    end
+  end
+
+  count
+end
+
+### 指数阶（循环实现）###
+def exponential(n)
+  count, base = 0, 1
+
+  # 细胞每轮一分为二，形成数列 1, 2, 4, 8, ..., 2^(n-1)
+  (0...n).each do
+    (0...base).each { count += 1 }
+    base *= 2
+  end
+
+  # count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
+  count
+end
+
+### 指数阶（递归实现）###
+def exp_recur(n)
+  return 1 if n == 1
+  exp_recur(n - 1) + exp_recur(n - 1) + 1
+end
+
+### 对数阶（循环实现）###
+def logarithmic(n)
+  count = 0
+
+  while n > 1
+    n /= 2
+    count += 1
+  end
+
+  count
+end
+
+### 对数阶（递归实现）###
+def log_recur(n)
+  return 0 unless n > 1
+  log_recur(n / 2) + 1
+end
+
+### 线性对数阶
+def linear_log_recur(n)
+  return 1 unless n > 1
+
+  count = linear_log_recur(n / 2) + linear_log_recur(n / 2)
+  (0...n).each { count += 1 }
+
+  count
+end
+
+### 阶乘阶（递归实现）###
+def factorial_recur(n)
+  return 1 if n == 0
+
+  count = 0
+  # 从 1 个分裂出 n 个
+  (0...n).each { count += factorial_recur(n - 1) }
+
+  count
+end
+
+### Driver Code ###
+
+# 可以修改 n 运行，体会一下各种复杂度的操作数量变化趋势
+n = 8
+puts "输入数据大小 n = #{n}"
+
+count = constant n
+puts "常数阶的操作数量 = #{count}"
+
+count = linear n
+puts "线性阶的操作数量 = #{count}"
+count = array_traversal Array.new n, 0
+puts "线性阶（遍历数组）的操作数量 = #{count}"
+
+count = quadratic n
+puts "平方阶的操作数量 = #{count}"
+nums = Array.new(n) { |i| n - i } # [n, n-1, ..., 2, 1]
+count = bubble_sort nums
+puts "平方阶（冒泡排序）的操作数量 = #{count}"
+
+count = exponential n
+puts "指数阶（循环实现）的操作数量 = #{count}"
+count = exp_recur n
+puts "指数阶（递归实现）的操作数量 = #{count}"
+
+count = logarithmic n
+puts "对数阶（循环实现）的操作数量 = #{count}"
+count = log_recur n
+puts "对数阶（递归实现）的操作数量 = #{count}"
+
+count = linear_log_recur n
+puts "线性对数阶（递归实现）的操作数量 = #{count}"
+
+count = factorial_recur n
+puts "阶乘阶（递归实现）的操作数量 = #{count}"

codes/ruby/chapter_computational_complexity/worst_best_time_complexity.rb

@@ -0,0 +1,34 @@
+=begin
+File: worst_best_time_complexity.rb
+Created Time: 2024-03-30
+Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
+=end
+
+### 生成一个数组，元素为: 1, 2, ..., n ，顺序被打乱 ###
+def random_number(n)
+  # 生成数组 nums =: 1, 2, 3, ..., n
+  nums = Array.new(n) { |i| i + 1 }
+  # 随机打乱数组元素
+  nums.shuffle!
+end
+
+### 查找数组 nums 中数字 1 所在索引 ###
+def find_one(nums)
+  for i in 0...nums.length
+    # 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
+    # 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
+    return i if nums[i] == 1
+  end
+
+  -1
+end
+
+### Driver Code ###
+
+for i in 0...10
+  n = 100
+  nums = random_number n
+  index = find_one nums
+  puts "\n数组 [ 1, 2, ..., n ] 被打乱后 = #{nums}"
+  puts "数字 1 的索引为 #{index}"
+end

codes/ruby/utils/print_util.rb

@@ -13,3 +13,46 @@ def print_linked_list(head)
   end
   puts "#{list.join(" -> ")}"
 end
+
+class Trunk
+  attr_accessor :prev, :str
+
+  def initialize(prev, str)
+    @prev = prev
+    @str = str
+  end
+end
+
+def show_trunk(p)
+  return if p.nil?
+
+  show_trunk(p.prev)
+  print p.str
+end
+
+### 打印二叉树 ###
+# This tree printer is borrowed from TECHIE DELIGHT
+# https://www.techiedelight.com/c-program-print-binary-tree/
+def print_tree(root, prev=nil, is_right=false)
+  return if root.nil?
+
+  prev_str = "    "
+  trunk = Trunk.new prev, prev_str
+  print_tree root.right, trunk, true
+
+  if prev.nil?
+    trunk.str = "———"
+  elsif is_right
+    trunk.str = "/———"
+    prev_str = "   |"
+  else
+    trunk.str = "\\———"
+    prev.str = prev_str
+  end
+
+  show_trunk trunk
+  puts " #{root.val}"
+  prev.str = prev_str if prev
+  trunk.str = "   |"
+  print_tree root.left, trunk, false
+end

codes/ruby/utils/tree_node.rb

@@ -0,0 +1,18 @@
+=begin
+File: tree_node.rb
+Created Time: 2024-03-30
+Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
+=end
+
+### 二叉树节点类 ###
+class TreeNode
+  attr_accessor :val    # 节点值
+  attr_accessor :height # 节点高度
+  attr_accessor :left   # 左子节点引用
+  attr_accessor :right  # 右子节点引用
+
+  def initialize(val=0)
+    @val = val
+    @height = 0
+  end
+end

docs/chapter_computational_complexity/space_complexity.md

@@ -335,7 +335,30 @@
 === "Ruby"
 
     ```ruby title=""
+    ### 类 ###
+    class Node
+        attr_accessor :val      # 节点值
+        attr_accessor :next     # 指向下一节点的引用
+
+        def initialize(x)
+            @val = x
+        end
+    end
+
+    ### 函数 ###
+    def function
+        # 执行某些操作...
+        0
+    end
 
+    ### 算法 ###
+    def algorithm(n)        # 输入数据
+        a = 0               # 暂存数据（常量）
+        b = 0               # 暂存数据（变量）
+        node = Node.new 0   # 暂存数据（对象）
+        c = function        # 栈帧空间（调用函数）
+        a + b + c           # 输出数据
+    end
     ```
 
 === "Zig"
@@ -499,7 +522,11 @@
 === "Ruby"
 
     ```ruby title=""
-
+    def algorithm(n)
+        a = 0                           # O(1)
+        b = Array.new 10000             # O(1)
+        nums = Array.new n if n > 10    # O(n)
+    end
     ```
 
 === "Zig"
@@ -763,7 +790,21 @@
 === "Ruby"
 
     ```ruby title=""
-
+    def function
+        # 执行某些操作
+        0
+    end
+
+    ### 循环的空间复杂度为 O(1) ###
+    def loop(n)
+        (0...n).each { function }
+    end
+
+    ### 递归的空间复杂度为 O(n) ###
+    def recur(n)
+        return if n == 1
+        recur n - 1
+    end
     ```
 
 === "Zig"

docs/chapter_computational_complexity/time_complexity.md

@@ -189,7 +189,16 @@
 === "Ruby"
 
     ```ruby title=""
-
+    # 在某运行平台下
+    def algorithm(n)
+        a = 2       # 1 ns 
+        a = a + 1   # 1 ns
+        a = a * 2   # 10 ns
+        # 循环 n 次
+        (n...0).each do # 1 ns
+            puts 0      # 5 ns
+        end
+    end
     ```
 
 === "Zig"
@@ -474,7 +483,20 @@ $$
 === "Ruby"
 
     ```ruby title=""
+    # 算法 A 的时间复杂度：常数阶
+    def algorithm_A(n)
+        puts 0
+    end
 
+    # 算法 B 的时间复杂度：线性阶
+    def algorithm_B(n)
+        (0...n).each { puts 0 }
+    end
+
+    # 算法 C 的时间复杂度：常数阶
+    def algorithm_C(n)
+        (0...1_000_000).each { puts 0 }
+    end
     ```
 
 === "Zig"
@@ -688,7 +710,15 @@ $$
 === "Ruby"
 
     ```ruby title=""
-
+    def algorithm(n)
+        a = 1       # +1
+        a = a + 1   # +1
+        a = a * 2   # +1
+        # 循环 n 次
+        (0...n).each do # +1
+            puts 0      # +1
+        end
+    end
     ```
 
 === "Zig"
@@ -970,7 +1000,16 @@ $T(n)$ 是一次函数，说明其运行时间的增长趋势是线性的，因
 === "Ruby"
 
     ```ruby title=""
-
+    def algorithm(n)
+        a = 1       # +0（技巧 1）
+        a = a + n   # +0（技巧 1）
+        # +n（技巧 2）
+        (0...(5 * n + 1)).each do { puts 0 }
+        # +n*n（技巧 3）
+        (0...(2 * n)).each do
+            (0...(n + 1)).each do { puts 0 }
+        end
+    end
     ```
 
 === "Zig"