将递归函数的返回值存储在ruby中的变量中(Storing the return value of a recursive function in a variable in ruby)

 我正在研究一个问题，找到一个数组中第一对与特定目标相加的数字。 我必须返回最右边的索引首先出现的那对。 所以说目标是10 ，发现的对是[5,5]在指数1,5和[3,7]在指数3,4我必须返回[3,7] 。 我有递归部分工作，问题是它返回一个散列，最左边的索引作为键，而对作为值。 我试图从另一个函数（#sum_pairs）调用这个递归函数，这样我就可以格式化我需要的输出，但是当我这样做并尝试将它存储在变量对中时，它返回的是一个空哈希，甚至虽然它只是在我从控制台调用它时返回完成的哈希。 
 
 下面是我的代码： 
 
    $starts = [0]

    def sum_pairs(ints, s)
     pairs = find_pairs(ints,s)
     pairs[pairs.keys.min]
    end

    def find_pairs(ints, s)
     sums = {}

     ints.each_with_index do |n, idx|
       next if idx < $starts.last
       right_idx = completes(ints, idx, n, s)

       if right_idx.nil?
        return sums if idx == (ints.length - 1)
        next
      else
       sums[right_idx] = [n, (s - n)]
       $starts << idx
       sums.merge!(sum_pairs(ints[0...right_idx], s))
       sums
      end
     end
    end

    def completes(ints, idx, n, s )
     ints.each_with_index do |num, index|
      next if index <= idx
      return index if num == s - n
     end
     nil
   end
 
 我知道我可以通过迭代来做到这一点，它可能会更有效或更高效，但我正在尝试使用它来更好地理解递归，谢谢！ 

I'm working on a problem to find the first pair of numbers in an array that sums to a specific target. I have to return the pair which its rightmost index appears first. So say target is 10 and pairs found are [5,5] at indices 1,5 and [3,7] at indices 3,4 I'd have to return [3,7]. I have the recursive part working, problem is it returns a hash with the leftmost index as key and the pair as value. I'm trying to call this recursive function from another function (#sum_pairs) so that I can then format the output I need, but when I do that and try to store it in the variable pairs all it returns is an empty hash, even though it returns the completed hash when I just call it from the console.
 
Heres my code: 
 
    $starts = [0]

    def sum_pairs(ints, s)
     pairs = find_pairs(ints,s)
     pairs[pairs.keys.min]
    end

    def find_pairs(ints, s)
     sums = {}

     ints.each_with_index do |n, idx|
       next if idx < $starts.last
       right_idx = completes(ints, idx, n, s)

       if right_idx.nil?
        return sums if idx == (ints.length - 1)
        next
      else
       sums[right_idx] = [n, (s - n)]
       $starts << idx
       sums.merge!(sum_pairs(ints[0...right_idx], s))
       sums
      end
     end
    end

    def completes(ints, idx, n, s )
     ints.each_with_index do |num, index|
      next if index <= idx
      return index if num == s - n
     end
     nil
   end
 
I know I could do this with iteration an it would probably be more or equally efficient, but I'm trying to use this to better understand recursion, thank you!

原文：https://stackoverflow.com/questions/39419518