how would one merge two strings that are ordered a

问题:

I am learning Lisp. I have implemented a common lisp function that merge two strings that are ordered alphabetically using recursion. Here is my code, but there is something wrong and I didn't figure it out.

(defun merge (F L)
    (if (null F)
        (if (null L)
            F         ; return f
            ( L ))    ; else return L
        ;else if
        (if (null L)
            F)        ; return F
    ;else if
    (if (string< (substring F 0 1) (substring L 0 1)
        (concat 'string (substring F 0 1) (merge (substring F 1 (length F)) L)))
    ( 
        (concat 'string (substring L 0 1) (merge F (substring L 1 (length L)) ))
    ))))

Edit : I simply want to merge two strings such as; inputs are string a = adf and string b = beg and the result or output should be; abdefg

Thanks in advance.

回答1:

Judging by your comments, it looks like you're trying to use if with a series of conditions (like a series of else ifs in some other languages). For that, you probably want cond.

I replaced that if with cond and cleaned up some other errors, and it worked.

(defun empty (s) (= (length s) 0))

(defun my-merge (F L)
  (cond 
   ((empty F)
    (if (empty L)
      F 
      L)) 
   ((empty L)
    F)
   (t
    (if (string< (subseq F 0 1) (subseq L 0 1))
      (concatenate 'string (subseq F 0 1) (my-merge (subseq F 1 (length F)) L)) 
      (concatenate 'string (subseq L 0 1) (my-merge F (subseq L 1 (length L))))))))

Your test case came out as you wanted it to:

* (my-merge "adf" "beg")

"abdefg"

回答2:

Using string< is an overkill, char< should be used instead, as shown by Kaz. Recalculating length at each step would make this algorithm quadratic, so should be avoided. Using sort to "fake it" makes it O(n log n) instead of O(n). Using concatenate 'string all the time probably incurs extra costs of unneeded traversals too.

Here's a natural recursive solution:

(defun str-merge (F L)
  (labels ((g (a b)
             (cond
               ((null a) b)
               ((null b) a)
               ((char< (car b) (car a))
                  (cons (car b) (g a (cdr b))))
               (t (cons (car a) (g (cdr a) b))))))
    (coerce (g (coerce F 'list) (coerce L 'list)) 'string)))

But, Common Lisp does not have a tail call optimization guarantee, let alone tail recursion modulo cons optimization guarantee (even if the latter was described as early as 1974, using "Lisp 1.6's rplaca and rplacd field assignment operators"). So we must hand-code this as a loop:

(defun str-merge (F L &aux (s (list nil)) )
  (do ( (p s (cdr p))
        (a (coerce F 'list) (if q a (cdr a)))
        (b (coerce L 'list) (if q (cdr b) b))
        (q nil))
      ( (or (null a) (null b))
          (if a (rplacd p a) (rplacd p b))
          (coerce (cdr s) 'string))
    (setq q (char< (car b) (car a)))
    (if q
      (rplacd p (list (car b)))
      (rplacd p (list (car a))))))

回答3:

There were quite a few good answers, so why would I add one more? Well, the below is probably more efficient then the other answers here.

(defun merge-strings (a b)
  (let* ((lena (length a))
         (lenb (length b))
         (len (+ lena lenb))
         (s (make-string len)))
    (labels
        ((safe-char< (x y)
           (if (and x y) (char< x y)
               (not (null x))))
         (choose-next (x y)
           (let ((ax (when (< x lena) (aref a x)))
                 (by (when (< y lenb) (aref b y)))
                 (xy (+ x y)))
             (cond
               ((= xy len) s)
               ((safe-char< ax by)
                (setf (aref s xy) ax)
                (choose-next (1+ x) y))
               (t
                (setf (aref s xy) by)
                (choose-next x (1+ y)))))))
      (choose-next 0 0))))

(merge-strings "adf" "beg")

It is more efficient specifically in the sense of memory allocations - it only allocates enough memory to write the result string, never coerces anything (from list to string or from array to string etc.) It may not look very pretty, but this is because it is trying to do every calculation only once.

This is, of course, not the most efficient way to write this function, but programming absolutely w/o efficiency in mind is not going to get you far.

回答4:

A recursive way to do it (fixed according to comment- other solutions can get an IF form as well).

(defun merge-strings (a b)
  (concatenate 'string
               (merge-strings-under a b)))

(defun merge-strings-under (a b)
  (when (and
       (= (length a)
          (length b))
       (> (length a) 0))
    (append (if (string< (aref a 0) (aref b 0))
                (list (aref a 0) (aref b 0))
                (list (aref b 0) (aref a 0)))
            (merge-strings-under (subseq a 1)
                           (subseq b 1)))))

---

Here's a iterative way to do it.

(concatenate 'string 
    (loop for i across "adf" for j across "beg" nconc (list i j)))

Note that these rely on building the string into a list of characters, then vectorizing it ( a string is a vector of characters).

You can also write a more C-esque approach...

(defun merge-strings-vector (a b)
  (let ((retstr (make-array (list (+
                                   (length a)
                                   (length b)))
                            :element-type 'character)))
    (labels ((merge-str (a b i)
               (when (and
                    (= (length a)
                       (length b))
                    (/= i (length a)))
                 (setf (aref retstr (* 2 i)) (aref a i))
                 (setf (aref retstr (1+ (* 2 i))) (aref b i))
                 (merge-str a b (1+ i)))))

      (merge-str a b 0)
      retstr)))

Note that this one - unlike the other 2 - has side effects within the function. It also, imo, is more difficult to understand.

All 3 take varying numbers of cycles to execute on SBCL 56; each seems to take between 6K and 11K on most of my trials. I'm not sure why.