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/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}"