各种排序的Ruby实现 博客分类： Ruby RubyJ#performance

Θ(n^2)

1, Bubble sort

def bubble_sort(a)
  (a.size-2).downto(0) do |i|
    (0..i).each do |j|
      a[j], a[j+1] = a[j+1], a[j] if a[j] > a[j+1]
    end
  end
  return a
end

2, Selection sort

def selection_sort(a)
  b = []
  a.size.times do |i|
    min = a.min
    b << min
    a.delete_at(a.index(min))
  end
  return b
end

3, Insertion sort

def insertion_sort(a)
  a.each_with_index do |el,i|
    j = i - 1
      while j >= 0
        break if a[j] <= el
        a[j + 1] = a[j]
        j -= 1
      end
    a[j + 1] = el
  end
  return a
end

4, Shell sort

def shell_sort(a)
  gap = a.size
  while(gap > 1)
    gap = gap / 2
    (gap..a.size-1).each do |i|
      j = i
      while(j > 0)
        a[j], a[j-gap] = a[j-gap], a[j] if a[j] <= a[j-gap]
        j = j - gap
      end
    end
  end
  return a
end

Θ(n*logn)

1, Merge sort

def merge(l, r)
  result = []
  while l.size > 0 and r.size > 0 do
    if l.first < r.first
      result << l.shift
    else
      result << r.shift
    end
  end
  if l.size > 0
    result += l
  end
  if r.size > 0
    result += r
  end
  return result
end

def merge_sort(a)
  return a if a.size <= 1
  middle = a.size / 2
  left = merge_sort(a[0, middle])
  right = merge_sort(a[middle, a.size - middle])
  merge(left, right)
end

2, Heap sort

def heapify(a, idx, size)
  left_idx = 2 * idx + 1
  right_idx = 2 * idx + 2
  bigger_idx = idx
  bigger_idx = left_idx if left_idx < size && a[left_idx] > a[idx]
  bigger_idx = right_idx if right_idx < size && a[right_idx] > a[bigger_idx]
  if bigger_idx != idx
    a[idx], a[bigger_idx] = a[bigger_idx], a[idx]
    heapify(a, bigger_idx, size)
  end
end

def build_heap(a)
  last_parent_idx = a.length / 2 - 1
  i = last_parent_idx
  while i >= 0
    heapify(a, i, a.size)
    i = i - 1
  end
end

def heap_sort(a)
  return a if a.size <= 1
  size = a.size
  build_heap(a)
  while size > 0
    a[0], a[size-1] = a[size-1], a[0]
    size = size - 1
    heapify(a, 0, size)
  end
  return a
end

3, Quick sort

def quick_sort(a)
  (x=a.pop) ? quick_sort(a.select{|i| i <= x}) + [x] + quick_sort(a.select{|i| i > x}) : []
end

Θ(n)

1, Counting sort

def counting_sort(a)
  min = a.min
  max = a.max
  counts = Array.new(max-min+1, 0)

  a.each do |n|
    counts[n-min] += 1
  end

  (0...counts.size).map{|i| [i+min]*counts[i]}.flatten
end

2, Radix sort

def kth_digit(n, i)
  while(i > 1)
    n = n / 10
    i = i - 1
  end
  n % 10
end

def radix_sort(a)
  max = a.max
  d = Math.log10(max).floor + 1

  (1..d).each do |i|
    tmp = []
    (0..9).each do |j|
      tmp[j] = []
    end

    a.each do |n|
      kth = kth_digit(n, i)
      tmp[kth] << n
    end
    a = tmp.flatten
  end
  return a
end

3, Bucket sort

def quick_sort(a)
  (x=a.pop) ? quick_sort(a.select{|i| i <= x}) + [x] + quick_sort(a.select{|i| i > x}) : []
end

def first_number(n)
  (n * 10).to_i
end

def bucket_sort(a)
  tmp = []
  (0..9).each do |j|
    tmp[j] = []
  end
  
  a.each do |n|
    k = first_number(n)
    tmp[k] << n
  end

  (0..9).each do |j|
    tmp[j] = quick_sort(tmp[j])
  end

  tmp.flatten
end

a = [0.75, 0.13, 0, 0.44, 0.55, 0.01, 0.98, 0.1234567]
p bucket_sort(a)

# Result: 
[0, 0.01, 0.1234567, 0.13, 0.44, 0.55, 0.75, 0.98]