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1

This question already has an answer here:

I am trying to write a method that takes an array of strings as its argument and returns array.reverse without actually using reverse (using Ruby). Most of my attempts have resulted in adding nils to the array. I can't track this algorithm down. Do you have a solution?

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marked as duplicate by Community♦ Jun 24 '15 at 3:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2  
Show one of those attempts! – Stan Jun 24 '15 at 3:09
up vote -1 down vote accepted

A pretty straightforward way is as follows:

def reverser(array)
  reversed = Array.new
  array.length.times { reversed << array.pop }
  reversed
end

We just pop off the last element of your array and push it into a new array until we've done so with all elements of your array.

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It should be noted that this mutates array. (Downvote not mine.) – Cary Swoveland Jun 24 '15 at 7:57
    
In addition to that, it is essentially the same as Cary's answer (except that this answer is inferior than Cary's). – sawa Jun 24 '15 at 19:40
    
I clicked this one because I am very new to coding. This seems the most straight ahead. I'm sure Cary's answer is better, but I am not yet familiar many of the methods and syntax. – Garett Arrowood Jun 24 '15 at 20:12
up vote 2 down vote

Many ways to do that. Here's one:

a = (1..10).to_a
  #=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

b = a.dup
a.size.times.with_object([]) { |_,c| c << b.pop }
  #=> [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
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up vote 1 down vote 
def reverse a; a.inject([], :unshift) end

reverse(["a", "b", "c"]) # => ["c", "b", "a"]
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