一些冗长的方法来做到这一点：

class Array
  def same_values?
    self.uniq.length == 1
  end
end

def find_boomerangs(arr)
  split_amount = 3 # count of a possible boomerang
  scan_amount = split_amount - 1 # scanning by index for array
  new_arr = []
  arr.each_with_index { |_, indx| # we only care for the indx
    end_of_indx = indx + scan_amount
    arry = arr[indx .. end_of_indx] # collect new possible boomerang from array
    next unless arry.count == split_amount # only check if possible boomerang is length of three

    new_arr << arry if arry.values_at(0, -1).same_values? && !arry.values_at(0, 1).same_values? # checks that the values at the start and end are the same and that the middle value is not the same as the first
  }
  new_arr # holds the boomerangs | call .count if you want the count
end

非常简单的方法（抱歉，用Python）：

def boomerangs(l):
    count = 0
    for i in range(len(l) - 2):
        if l[i] == l[i+2] and l[i] != l[i+1]:
            count += 1
    return count

print(boomerangs([3, 7, 3, 2, 1, 5, 1, 2, 2, -2, 2]))
print(boomerangs([1, 7, 1, 7, 1, 7, 1]))

def count_boomerangs(arr)
  arr.each_cons(3).count { |a,b,c| a == c && a !=b }
end

count_boomerangs [3, 7, 3, 2, 1, 5, 1, 2, 2, -2, 2] #=> 3
count_boomerangs [1, 7, 1, 7, 1, 7, 1]              #=> 5
count_boomerangs [9, 5, 9, 5, 1, 1, 1]              #=> 2
count_boomerangs [5, 6, 6, 7, 6, 3, 9]              #=> 1
count_boomerangs [4, 4, 4, 9, 9, 9, 9]              #=> 0

请参见可枚举的#each_cons。

get_bommerangs = lambda arr: sum(
    arr[i - 1] == arr[i + 1] and arr[i - 1] != arr[i] for i in range(1, len(arr) - 1)
)

bommerangs = get_bommerangs([2, 1, 2, 3, 3, 4, 2, 4])