Sorting an Array in Ruby without using Sort method

Question

I'm trying to use the bubble sort method to sort an array of only three numbers. The code I'm using is below.

def my_sort(list)
  return list if list.size <= 1 

  swapped = false

  while !swapped
    swapped = false

    0.upto(list.size-2) do |i|
      if list[i] > list[i+1]
        list[i], list[i+1] = list[i+1], list[i]
        swapped = true
      end
    end

    list
  end

my_sort([3,1,2])

Here is the error message I keep getting:

Syntax error, unexpected $end, expecting keyword_end

I was just wondering which end shouldn't be included?

Indent the code properly and you'll see the missing end right away. — tokland
– tokland, Commented Jun 19, 2012 at 18:26
possible duplicate of How do you sort without using the sort method? — Emil Vikström
– Emil Vikström, Commented Jun 24, 2012 at 22:11

rfunduk · Accepted Answer · 2012-06-19 18:27:33Z

3

You're missing an end after swapped = true. It would be best to indent your code thoroughly and accurately to avoid this kind of problem:

def my_sort(list)
  return list if list.size <= 1 

  swapped = false
  while !swapped
    swapped = false
    0.upto(list.size-2) do |i|
      if list[i] > list[i+1]
        list[i], list[i+1] = list[i+1], list[i]
        swapped = true
      end
    end
  end

  list
end

answered Jun 19, 2012 at 18:27

rfunduk

30.5k5 gold badges61 silver badges55 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Vieenay Siingh Over a year ago

this did not work. list = [14,11,36,42,12,10] is returning sorted array as [11, 14, 36, 12, 10, 42] which is wrong.

Pablo Over a year ago

Your solution doesn't work; it returned the same array unsorted.

rfunduk Over a year ago

this is the OPs code, but correctly indented and with the syntax error fixed

Kyle · Accepted Answer · 2012-06-19 18:27:03Z

3

You're missing an end

  if list[i] > list[i+1]
    list[i], list[i+1] = list[i+1], list[i]
    swapped = true
  end # <--------------------------------------------------------

Edit: As the other answer mentions, indent your code to make these errors more visible.

answered Jun 19, 2012 at 18:27

Kyle

22.3k2 gold badges63 silver badges63 bronze badges

Comments

V K Singh · Accepted Answer · 2019-11-09 22:07:29Z

2

This looks a better and quicker one.

arr = [1,5,7,2,3,50,78,34, 1, 15, 89, 8]

def sort_array(arr)
  arr.size.times do
    arr.map.with_index do |num, index|
      next if index.eql?(arr.size - 1)
      arr[index], arr[index+1] = arr[index+1], arr[index] if num > arr[index+1]
    end
  end
end

Call the above method as print sort_array(arr) to get expected result.

answered Nov 9, 2019 at 22:07

V K Singh

1,25411 silver badges14 bronze badges

Comments

Melissa Quintero · Accepted Answer · 2016-07-10 22:41:39Z

1

Your code works for that specific array. Because your loop is looking if the next element is higher, then swipe. But what about more elements in the array? This is recursive solution for all cases.

def my_sort(list, new_array = nil)

  return new_array if list.size <= 0
  if new_array == nil
    new_array = []
  end
  min = list.min
  new_array << min
  list.delete(min)

  my_sort(list, new_array)

end

puts my_sort([3, 1, 2, 20, 11, 14, 3, 6, 8, 5])

answered Jul 10, 2016 at 22:41

Melissa Quintero

1911 silver badge4 bronze badges

Comments

Vieenay Siingh · Accepted Answer · 2019-04-04 10:30:41Z

0

What worked for me, is below.

def my_sort(list)   
  n = list.length   
  loop do
    swapped = false
    (n-1).times do |i|
      if list[i] > list[i+1]
        list[i], list[i+1] = list[i+1], list[i]
        swapped = true
      end
    end
    break if not swapped  
  end   
  list 
end

answered Apr 4, 2019 at 10:30

Vieenay Siingh

8672 gold badges19 silver badges45 bronze badges

Comments

Adrian Mole · Accepted Answer · 2020-11-19 11:20:01Z

0

def sort(arr)
    arr_len = arr.length - 1
    swap = true
    while swap
        swap = false
        arr_len.times do |i|
         if arr[i] > arr[i+1]
           arr[i],arr[i + 1] = arr[i + 1],arr[i]
           swap = true
         end
        end
     end
     p arr

end

edited Nov 19, 2020 at 11:20

Adrian Mole

52.1k193 gold badges61 silver badges101 bronze badges

answered Nov 19, 2020 at 11:18

Anup Baidya

1

1 Comment

frianH Over a year ago

Please add a bit explanation why this code should work.

Ashwin · Accepted Answer · 2022-05-04 14:03:01Z

0

arr = [1,2,8,4,5,3,1,0,6]

0.upto(arr.size - 1) do |i|
  (i+1).upto(arr.size - 1) do |j|
    if arr[i] > arr[j]
      arr[j],arr[i] = arr[i],arr[j]
    end
  end
end

answered May 4, 2022 at 14:03

Ashwin

116 bronze badges

Comments

Ganesh Kaliannan · Accepted Answer · 2017-09-15 00:17:01Z

-1

#Using bubble sort algorithm in ruby

a = [1,5,7,2,3,50,78,34,89]

a.size.times.each do |t|
 i=0
 a.each do |b|
   if b > a[i+1]
     a[i],a[i+1] = a[i+1],a[i]
   end
   i+=1 if i < a.size-2
 end
end
print a
#output: [1, 2, 3, 5, 7, 34, 50, 78, 89]

answered Sep 15, 2017 at 0:17

Ganesh Kaliannan

11 bronze badge

1 Comment

FluffyKitten Over a year ago

Code-only answers are discouraged because they do not explain how they resolve the issue. Please update your answer to explain how this improves on the other accepted and upvoted answers this question already has. Also, this question is 5 years old, your efforts would be more appreciated by users who have recent unanswered questions. Please review How do I write a good answer.

Collectives™ on Stack Overflow

Sorting an Array in Ruby without using Sort method

8 Answers 8

3 Comments

Comments

Comments

Comments

Comments

1 Comment

Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

8 Answers 8

3 Comments

Comments

Comments

Comments

Comments

1 Comment

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related