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?

Answer 1

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.

Answer 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.

Answer 3

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])

Answer 4

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   

Answer 5

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

Answer 6

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

Answer 7

#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]