Ruby各种排序算法的实现方法
我们经常需要在ruby中使用排序算法,但是在ruby中排序算法有很多我们要怎么才能实现呢?那么接下来的内容中我们就一起去看看Ruby各种排序算法的实现方法。
时间复杂度:Θ(n^2)
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
Selection sort
def selection_sort(a)
b = []
a.size.times do |i|
min = a.min
b a.delete_at(a.index(min))
end
return b
end
Insertion sort
def insertion_sort(a)
a.each_with_index do |el,i|
j = i - 1
while j >= 0
break if a[j] a[j + 1] = a[j]
j -= 1
end
a[j + 1] = el
end
return a
end
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] j = j - gap
end
end
end
return a
end
时间复杂度:Θ(n*logn)
Merge sort
def merge(l, r)
result = []
while l.size > 0 and r.size > 0 do
if l.first result else
result 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 middle = a.size / 2
left = merge_sort(a[0, middle])
right = merge_sort(a[middle, a.size - middle])
merge(left, right)
end
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 a[idx]
bigger_idx = right_idx if 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
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
Quick sort
def quick_sort(a)
(x=a.pop) ? quick_sort(a.select{|i| i x}) : []
end
时间复杂度:Θ(n)
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
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] end
a = tmp.flatten
end
return a
end
Bucket sort
def quick_sort(a)
(x=a.pop) ? 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] 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]
上文就是小编介绍Ruby各种排序算法的实现方法,希望这篇文章能够给大家的学习工作带来帮助,知识水平有限文章中如有错误还请批评指正!