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* Bug fixes * Fix the term in heap figures * Unify the font of the chapter covers for the zh, en, and zh-Hant version * Sync the zh-hant vertion with the main branch * Update README for testing * Update README for testing * Update README for testing * Update README for zh, en, zh-hant version * Fix the issue links * Update README * Update README * edition -> version
164 lines
3.2 KiB
Ruby
164 lines
3.2 KiB
Ruby
=begin
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File: time_complexity.rb
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Created Time: 2024-03-30
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 常数阶 ###
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def constant(n)
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count = 0
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size = 100000
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(0...size).each { count += 1 }
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count
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end
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### 线性阶 ###
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def linear(n)
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count = 0
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(0...n).each { count += 1 }
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count
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end
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### 线性阶(遍历数组)###
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def array_traversal(nums)
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count = 0
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# 循环次数与数组长度成正比
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for num in nums
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count += 1
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end
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count
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end
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### 平方阶 ###
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def quadratic(n)
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count = 0
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# 循环次数与数据大小 n 成平方关系
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for i in 0...n
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for j in 0...n
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count += 1
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end
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end
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count
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end
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### 平方阶(冒泡排序)###
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def bubble_sort(nums)
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count = 0 # 计数器
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# 外循环:未排序区间为 [0, i]
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for i in (nums.length - 1).downto(0)
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# 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
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for j in 0...i
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if nums[j] > nums[j + 1]
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# 交换 nums[j] 与 nums[j + 1]
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tmp = nums[j]
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nums[j] = nums[j + 1]
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nums[j + 1] = tmp
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count += 3 # 元素交换包含 3 个单元操作
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end
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end
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end
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count
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end
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### 指数阶(循环实现)###
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def exponential(n)
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count, base = 0, 1
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# 细胞每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
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(0...n).each do
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(0...base).each { count += 1 }
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base *= 2
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end
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# count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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count
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end
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### 指数阶(递归实现)###
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def exp_recur(n)
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return 1 if n == 1
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exp_recur(n - 1) + exp_recur(n - 1) + 1
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end
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### 对数阶(循环实现)###
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def logarithmic(n)
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count = 0
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while n > 1
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n /= 2
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count += 1
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end
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count
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end
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### 对数阶(递归实现)###
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def log_recur(n)
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return 0 unless n > 1
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log_recur(n / 2) + 1
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end
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### 线性对数阶 ###
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def linear_log_recur(n)
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return 1 unless n > 1
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count = linear_log_recur(n / 2) + linear_log_recur(n / 2)
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(0...n).each { count += 1 }
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count
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end
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### 阶乘阶(递归实现)###
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def factorial_recur(n)
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return 1 if n == 0
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count = 0
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# 从 1 个分裂出 n 个
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(0...n).each { count += factorial_recur(n - 1) }
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count
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end
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### Driver Code ###
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# 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
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n = 8
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puts "输入数据大小 n = #{n}"
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count = constant(n)
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puts "常数阶的操作数量 = #{count}"
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count = linear(n)
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puts "线性阶的操作数量 = #{count}"
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count = array_traversal(Array.new(n, 0))
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puts "线性阶(遍历数组)的操作数量 = #{count}"
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count = quadratic(n)
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puts "平方阶的操作数量 = #{count}"
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nums = Array.new(n) { |i| n - i } # [n, n-1, ..., 2, 1]
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count = bubble_sort(nums)
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puts "平方阶(冒泡排序)的操作数量 = #{count}"
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count = exponential(n)
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puts "指数阶(循环实现)的操作数量 = #{count}"
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count = exp_recur(n)
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puts "指数阶(递归实现)的操作数量 = #{count}"
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count = logarithmic(n)
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puts "对数阶(循环实现)的操作数量 = #{count}"
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count = log_recur(n)
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puts "对数阶(递归实现)的操作数量 = #{count}"
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count = linear_log_recur(n)
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puts "线性对数阶(递归实现)的操作数量 = #{count}"
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count = factorial_recur(n)
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puts "阶乘阶(递归实现)的操作数量 = #{count}"
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