Problem 92

*4clojure-problem-92*.clj

;; 4Clojure Question 92
;;
;; Roman numerals are easy to recognize, but not everyone knows all the rules necessary to work with them. Write a function to parse a Roman-numeral string and return the number it represents.
;;
;; <br /><br />
;;
;; You can assume that the input will be well-formed, in upper-case, and follow the <a href="http://en.wikipedia.org/wiki/Roman_numerals#Subtractive_principle">subtractive principle</a>. You don't need to handle any numbers greater than MMMCMXCIX (3999), the largest number representable with ordinary letters.
;;
;; Use M-x 4clojure-check-answers when you're done!

(= 14 (__ "XIV"))

(= 827 (__ "DCCCXXVII"))

(= 3999 (__ "MMMCMXCIX"))

(= 48 (__ "XLVIII"))

;; mine:
(fn [letters]
  (let [order (zipmap "IVXLCDM" [1 5 10 50 100 500 1000])
        cmp   (fn [a b] (compare (get order a) (get order b)))
        magn  (fn [[letter :as g]] (* (get order letter) (count g)))]
    (loop [[f s :as xs] (partition-by identity letters)
           acc 0]
      (if-not s
        (+ (magn f) acc)
        (recur (rest xs)
               (-> (magn f)
                   (* (cmp (first f) (first s)))
                   (+ acc)))))))

;; austintaylor's solution:
(fn [s]
  (let [roman {"M" 1000 "CM" 900 "D" 500 "CD" 400
    "C" 100 "XC" 90 "L" 50 "XL" 40 "X" 10 "IX" 9
    "V" 5 "IV" 4 "I" 1}]
    (reduce + (map roman
      (re-seq #"CM|CD|XC|XL|IX|IV|[MDCLXVI]" s)))))

bbb.clj

(fn [letters]
  (let [order  (zipmap "IVXLCDM" [1 5 10 50 100 500 1000])
        cmp    (fn [a b] (compare (get order a) (get order b)))
        magn   (fn [[letter :as g]] (* (get order letter) (count g)))
        groups (partition-by identity letters)]
    (+ (magn (last groups))
       (reduce
        (fn [acc [[f :as fs] [s]]]
          (-> (magn fs)
              (* (cmp f s))
              (+ acc)))
        0 (partition 2 1 groups)))))