chore: update driver code (#1265)

codes/ruby/chapter_computational_complexity/iteration.rb

@@ -62,17 +62,18 @@ def nested_for_loop(n)
 end
 
 ### Driver Code ###
+if __FILE__ == $0
+  n = 5
 
-n = 5
+  res = for_loop(n)
+  puts "\nfor 循环的求和结果 res = #{res}"
 
-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(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 = 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}"
+  res = nested_for_loop(n)
+  puts "\n双层 for 循环的遍历结果 #{res}"
+end

codes/ruby/chapter_computational_complexity/recursion.rb

@@ -53,17 +53,18 @@ def fib(n)
 end
 
 ### Driver Code ###
+if __FILE__ == $0
+  n = 5
 
-n = 5
+  res = recur(n)
+  puts "\n递归函数的求和结果 res = #{res}"
 
-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 = 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 = 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}"
+  res = fib(n)
+  puts "\n斐波那契数列的第 #{n} 项为 #{res}"
+end

codes/ruby/chapter_computational_complexity/space_complexity.rb

@@ -72,20 +72,21 @@ def build_tree(n)
 end
 
 ### Driver Code ###
+if __FILE__ == $0
+  n = 5
 
-n = 5
+  # 常数阶
+  constant(n)
 
-# 常数阶
-constant(n)
+  # 线性阶
+  linear(n)
+  linear_recur(n)
 
-# 线性阶
-linear(n)
-linear_recur(n)
+  # 平方阶
+  quadratic(n)
+  quadratic_recur(n)
 
-# 平方阶
-quadratic(n)
-quadratic_recur(n)
-
-# 指数阶
-root = build_tree(n)
-print_tree(root)
+  # 指数阶
+  root = build_tree(n)
+  print_tree(root)
+end

codes/ruby/chapter_computational_complexity/time_complexity.rb

@@ -128,37 +128,38 @@ def factorial_recur(n)
 end
 
 ### Driver Code ###
+if __FILE__ == $0
+  # 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}"
 
-# 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 = 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 = 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 = 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 = 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 = 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 = 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}"
+  count = factorial_recur(n)
+  puts "阶乘阶（递归实现）的操作数量 = #{count}"
+end

codes/ruby/chapter_computational_complexity/worst_best_time_complexity.rb

@@ -24,11 +24,12 @@ def find_one(nums)
 end
 
 ### Driver Code ###
-
-for i in 0...10
-  n = 100
-  nums = random_numbers(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}"
+if __FILE__ == $0
+  for i in 0...10
+    n = 100
+    nums = random_numbers(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
 end