我想写一个函数,它以一个字母数组作为参数,并从中选择一些字母。

假设您提供了一个包含8个字母的数组,并希望从中选择3个字母。那么你应该得到:

8! / ((8 - 3)! * 3!) = 56

数组(或单词),每个数组由3个字母组成。


当前回答

说了这么多,做了这么多,这就是奥卡姆的代码。 算法是显而易见的代码..

let combi n lst =
    let rec comb l c =
        if( List.length c = n) then [c] else
        match l with
        [] -> []
        | (h::t) -> (combi t (h::c))@(combi t c)
    in
        combi lst []
;;

其他回答

也许我错过了重点(你需要的是算法,而不是现成的解决方案),但看起来scala已经开箱即用了(现在):

def combis(str:String, k:Int):Array[String] = {
  str.combinations(k).toArray 
}

使用这样的方法:

  println(combis("abcd",2).toList)

会产生:

  List(ab, ac, ad, bc, bd, cd)
static IEnumerable<string> Combinations(List<string> characters, int length)
{
    for (int i = 0; i < characters.Count; i++)
    {
        // only want 1 character, just return this one
        if (length == 1)
            yield return characters[i];

        // want more than one character, return this one plus all combinations one shorter
        // only use characters after the current one for the rest of the combinations
        else
            foreach (string next in Combinations(characters.GetRange(i + 1, characters.Count - (i + 1)), length - 1))
                yield return characters[i] + next;
    }
}

Lisp宏为所有值r(每次取)生成代码

(defmacro txaat (some-list taken-at-a-time)
  (let* ((vars (reverse (truncate-list '(a b c d e f g h i j) taken-at-a-time))))
    `(
      ,@(loop for i below taken-at-a-time 
           for j in vars 
           with nested = nil 
           finally (return nested) 
           do
             (setf 
              nested 
              `(loop for ,j from
                    ,(if (< i (1- (length vars)))
                         `(1+ ,(nth (1+ i) vars))
                         0)
                  below (- (length ,some-list) ,i)
                    ,@(if (equal i 0) 
                          `(collect 
                               (list
                                ,@(loop for k from (1- taken-at-a-time) downto 0
                                     append `((nth ,(nth k vars) ,some-list)))))
                          `(append ,nested))))))))

So,

CL-USER> (macroexpand-1 '(txaat '(a b c d) 1))
(LOOP FOR A FROM 0 TO (- (LENGTH '(A B C D)) 1)
    COLLECT (LIST (NTH A '(A B C D))))
T
CL-USER> (macroexpand-1 '(txaat '(a b c d) 2))
(LOOP FOR A FROM 0 TO (- (LENGTH '(A B C D)) 2)
      APPEND (LOOP FOR B FROM (1+ A) TO (- (LENGTH '(A B C D)) 1)
                   COLLECT (LIST (NTH A '(A B C D)) (NTH B '(A B C D)))))
T
CL-USER> (macroexpand-1 '(txaat '(a b c d) 3))
(LOOP FOR A FROM 0 TO (- (LENGTH '(A B C D)) 3)
      APPEND (LOOP FOR B FROM (1+ A) TO (- (LENGTH '(A B C D)) 2)
                   APPEND (LOOP FOR C FROM (1+ B) TO (- (LENGTH '(A B C D)) 1)
                                COLLECT (LIST (NTH A '(A B C D))
                                              (NTH B '(A B C D))
                                              (NTH C '(A B C D))))))
T

CL-USER> 

And,

CL-USER> (txaat '(a b c d) 1)
((A) (B) (C) (D))
CL-USER> (txaat '(a b c d) 2)
((A B) (A C) (A D) (B C) (B D) (C D))
CL-USER> (txaat '(a b c d) 3)
((A B C) (A B D) (A C D) (B C D))
CL-USER> (txaat '(a b c d) 4)
((A B C D))
CL-USER> (txaat '(a b c d) 5)
NIL
CL-USER> (txaat '(a b c d) 0)
NIL
CL-USER> 

我的实现在c/c++

#include <unistd.h>
#include <stdio.h>
#include <iconv.h>
#include <string.h>
#include <errno.h>
#include <stdlib.h>

int main(int argc, char **argv)
{
    int opt = -1, min_len = 0, max_len = 0;
    char ofile[256], fchar[2], tchar[2];
    ofile[0] = 0;
    fchar[0] = 0;
    tchar[0] = 0;
    while((opt = getopt(argc, argv, "o:f:t:l:L:")) != -1)
    {
            switch(opt)
            {
                    case 'o':
                    strncpy(ofile, optarg, 255);
                    break;
                    case 'f':
                    strncpy(fchar, optarg, 1);
                    break;
                    case 't':
                    strncpy(tchar, optarg, 1);
                    break;
                    case 'l':
                    min_len = atoi(optarg);
                    break;
                    case 'L':
                    max_len = atoi(optarg);
                    break;
                    default:
                    printf("usage: %s -oftlL\n\t-o output file\n\t-f from char\n\t-t to char\n\t-l min seq len\n\t-L max seq len", argv[0]);
            }
    }
if(max_len < 1)
{
    printf("error, length must be more than 0\n");
    return 1;
}
if(min_len > max_len)
{
    printf("error, max length must be greater or equal min_length\n");
    return 1;
}
if((int)fchar[0] > (int)tchar[0])
{
    printf("error, invalid range specified\n");
    return 1;
}
FILE *out = fopen(ofile, "w");
if(!out)
{
    printf("failed to open input file with error: %s\n", strerror(errno));
    return 1;
}
int cur_len = min_len;
while(cur_len <= max_len)
{
    char buf[cur_len];
    for(int i = 0; i < cur_len; i++)
        buf[i] = fchar[0];
    fwrite(buf, cur_len, 1, out);
    fwrite("\n", 1, 1, out);
    while(buf[0] != (tchar[0]+1))
    {
        while(buf[cur_len-1] < tchar[0])
        {
            (int)buf[cur_len-1]++;
            fwrite(buf, cur_len, 1, out);
            fwrite("\n", 1, 1, out);
        }
        if(cur_len < 2)
            break;
        if(buf[0] == tchar[0])
        {
            bool stop = true;
            for(int i = 1; i < cur_len; i++)
            {
                if(buf[i] != tchar[0])
                {
                    stop = false;
                    break;
                }
            }
            if(stop)
                break;
        }
        int u = cur_len-2;
        for(; u>=0 && buf[u] >= tchar[0]; u--)
            ;
        (int)buf[u]++;
        for(int i = u+1; i < cur_len; i++)
            buf[i] = fchar[0];
        fwrite(buf, cur_len, 1, out);
        fwrite("\n", 1, 1, out);
    }
    cur_len++;
}
fclose(out);
return 0;
}

这里我的实现在c++,它写所有的组合到指定的文件,但行为可以改变,我在生成各种字典,它接受最小和最大长度和字符范围,目前只有ANSI支持,它足以满足我的需求

下面是我最近用Java写的一段代码,它计算并返回从“outOf”元素中“num”元素的所有组合。

// author: Sourabh Bhat (heySourabh@gmail.com)

public class Testing
{
    public static void main(String[] args)
    {

// Test case num = 5, outOf = 8.

        int num = 5;
        int outOf = 8;
        int[][] combinations = getCombinations(num, outOf);
        for (int i = 0; i < combinations.length; i++)
        {
            for (int j = 0; j < combinations[i].length; j++)
            {
                System.out.print(combinations[i][j] + " ");
            }
            System.out.println();
        }
    }

    private static int[][] getCombinations(int num, int outOf)
    {
        int possibilities = get_nCr(outOf, num);
        int[][] combinations = new int[possibilities][num];
        int arrayPointer = 0;

        int[] counter = new int[num];

        for (int i = 0; i < num; i++)
        {
            counter[i] = i;
        }
        breakLoop: while (true)
        {
            // Initializing part
            for (int i = 1; i < num; i++)
            {
                if (counter[i] >= outOf - (num - 1 - i))
                    counter[i] = counter[i - 1] + 1;
            }

            // Testing part
            for (int i = 0; i < num; i++)
            {
                if (counter[i] < outOf)
                {
                    continue;
                } else
                {
                    break breakLoop;
                }
            }

            // Innermost part
            combinations[arrayPointer] = counter.clone();
            arrayPointer++;

            // Incrementing part
            counter[num - 1]++;
            for (int i = num - 1; i >= 1; i--)
            {
                if (counter[i] >= outOf - (num - 1 - i))
                    counter[i - 1]++;
            }
        }

        return combinations;
    }

    private static int get_nCr(int n, int r)
    {
        if(r > n)
        {
            throw new ArithmeticException("r is greater then n");
        }
        long numerator = 1;
        long denominator = 1;
        for (int i = n; i >= r + 1; i--)
        {
            numerator *= i;
        }
        for (int i = 2; i <= n - r; i++)
        {
            denominator *= i;
        }

        return (int) (numerator / denominator);
    }
}