#26 fibonacci seqs

Difficulty: Easy

Topic: Fibonacci seqs

Write a function which returns the first X fibonacci numbers.

;; 首先实现一个求fibonacci数的函数
;;最简单的实现，就是通过定义来实现递归函数，（如下的fibonacci-number），但是这样缺点很明显，首先这不算尾递归，代码里面有大量的重复计算。其次，jvm不支持尾调用优化，因此，即使是尾递归，当嵌套层侧过深时，也会出现stackoverflow，
;; bad implementation!!!
(defn fibonacci-number [number]
  (cond
    (= 1 number) 1
    (= 2 number) 1
    :default (+ (fibonacci-number (- number 1))
                (fibonacci-number (- number 2)))))

;; also bad implementation!!!
(defn fib-num-helper [num result1 result2]
  (cond
    (= 1 num) result1
    (= 2 num) result2
    :default (fib-num-helper (dec num) result2 (+ result1 result2))))

;;test
(fib-num-helper 3 1 1)

;;not good
(defn fib-num [num]
  (let [result1 1
        result2 1]
    ((fn helper  [n r1 r2]
       (cond
         (= 1 n) r1
         (= 2 n) r2
         :default (helper (dec n) r2 (+ r1 r2))))
     num result1 result2)))
;;test
(fib-num 14)

(defn new-fib-num [num]
  (defn new-fib-num-helper [counter
                           limit
                           result1
                           result2
                           r-seq]
    (cond
      (= counter limit) (seq r-seq)
      :default (recur (inc counter)
                      limit
                      result2
                      (+ result1 result2)
                      (conj r-seq result2))))
  (new-fib-num-helper 1
                      num
                      1
                      1
                      [1]))
;;test
(new-fib-num 4)
(new-fib-num 8)