Sorting inputed numbers on array - Ruby - without using 'sort'

Question

I would really appreciate it if someone can help me on this. I want to input numbers on this array x[0,100] Then, print the whole array but with the numbers inside sorted from the smallest to biggest WITHOUT using sort array.

I have this code so far, but I get an error where I do the comparison. Can someone help me on this please?

x = [0,100]
print x 

puts "\nInput any number from 0~100"
num = gets.to_i
x.push(num)

for i in 0..x.length-1
    for j in 0..x.length-1
        if x[j] < x[j+1] 
            swap = x[j]
            x[j] = x[j+1]
            x[j+1] = swap
        end
    end
end
p x

Is there any good reason you don't want to use standard API ? WITHOUT using sort array — Jean Bob
– Jean Bob, Commented Oct 19, 2015 at 9:14
My professor doesn't want us to use sort. I think it is because he want us to think our way around — bisuke
– bisuke, Commented Oct 19, 2015 at 9:16
You can swap in one line like this:` x[j], x[j+1] = x[j+1], x[j]` — steenslag
– steenslag, Commented Oct 19, 2015 at 15:29

bisuke · Accepted Answer · 2015-10-19 12:33:40Z

5

Regardless of what the best solution would be, you get an error because your second loop is going out of bound: you're calling x[j+1] up to x.length-1, which means that you end up calling x[x.length] when the last element is x[x.length - 1].

Just replace "for j in 0..x.length-1" with "for j in 0..x.length-2"

x = [0,100]
print x

puts "\nInput any number from 0~100"
num = gets.to_i
x.push(num)

for i in 0..x.length-1
    for j in 0..x.length-2
        if x[j] < x[j+1]
            swap = x[j]
            x[j] = x[j+1]
            x[j+1] = swap
        end
    end
end
p x

edited Oct 19, 2015 at 12:33

bisuke

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answered Oct 19, 2015 at 9:18

Arnaud

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4 Comments

DaniG2k Over a year ago

Why are you doing x = [0,100] and x = [1,49,20,30,58,10,48,9] at the beginning of this script? Also, it's generally more Ruby-esque to use each instead of for loops :)

Arnaud Over a year ago

@DaniG2k Because that's what the OP wrote ;) I just copy/pasted the code of the OP with the correction, you should ask him/her

DaniG2k Over a year ago

Oops that comment was meant for @basukesu not for you :P

bisuke Over a year ago

oops! I am sorry. I was trying different things out before I posted my code, and I forgot to delete that useless array! Thank you for noticing @DaniG2k I just deleted it but it will take some time for my edit to be processed :/

Melissa Quintero · Accepted Answer · 2016-07-10 22:43:31Z

0

Ruby ranges behave pretty much like arrays, at least when looping. They are already in order.

(1..100).each { |num| puts num }

Your code looks like it is looking if the next element is higher, then swipe. But what about more elements in the array? This is recursive way

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])

edited Jul 10, 2016 at 22:43

answered Jul 10, 2016 at 22:10

Melissa Quintero

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