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
下面是c++中的迭代算法，它不使用STL，也不使用递归，也不使用条件嵌套循环。这样更快，它不执行任何元素交换，也不会给堆栈带来递归负担，还可以通过分别用mallloc()、free()和printf()替换new、delete和std::cout轻松地移植到ANSI C。
如果你想用不同或更长的字母显示元素，那么改变*字母参数以指向不同于"abcdefg"的字符串。
void OutputArrayChar(unsigned int* ka, size_t n, const char *alphabet) {
for (int i = 0; i < n; i++)
std::cout << alphabet[ka[i]] << ",";
std::cout << endl;
}
void GenCombinations(const unsigned int N, const unsigned int K, const char *alphabet) {
unsigned int *ka = new unsigned int [K]; //dynamically allocate an array of UINTs
unsigned int ki = K-1; //Point ki to the last elemet of the array
ka[ki] = N-1; //Prime the last elemet of the array.
while (true) {
unsigned int tmp = ka[ki]; //Optimization to prevent reading ka[ki] repeatedly
while (ki) //Fill to the left with consecutive descending values (blue squares)
ka[--ki] = --tmp;
OutputArrayChar(ka, K, alphabet);
while (--ka[ki] == ki) { //Decrement and check if the resulting value equals the index (bright green squares)
OutputArrayChar(ka, K, alphabet);
if (++ki == K) { //Exit condition (all of the values in the array are flush to the left)
delete[] ka;
return;
}
}
}
}
int main(int argc, char *argv[])
{
GenCombinations(7, 4, "abcdefg");
return 0;
}
重要提示:字母参数*必须指向至少N个字符的字符串。你也可以传递一个在其他地方定义的字符串地址。
组合:从“7选4”中选择。
其他回答
Python中的简短示例:
def comb(sofar, rest, n):
if n == 0:
print sofar
else:
for i in range(len(rest)):
comb(sofar + rest[i], rest[i+1:], n-1)
>>> comb("", "abcde", 3)
abc
abd
abe
acd
ace
ade
bcd
bce
bde
cde
为了解释，递归方法用下面的例子描述:
示例:A B C D E 3的所有组合是:
A与其余2的所有组合(B C D E) B与其余2的所有组合(C D E) C与其余2的所有组合(D E)
在c#中:
public static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> elements, int k)
{
return k == 0 ? new[] { new T[0] } :
elements.SelectMany((e, i) =>
elements.Skip(i + 1).Combinations(k - 1).Select(c => (new[] {e}).Concat(c)));
}
用法:
var result = Combinations(new[] { 1, 2, 3, 4, 5 }, 3);
结果:
123
124
125
134
135
145
234
235
245
345
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>
算法:
从1数到2^n。 将每个数字转换为二进制表示。 根据位置，将每个“on”位转换为集合中的元素。
在c#中:
void Main()
{
var set = new [] {"A", "B", "C", "D" }; //, "E", "F", "G", "H", "I", "J" };
var kElement = 2;
for(var i = 1; i < Math.Pow(2, set.Length); i++) {
var result = Convert.ToString(i, 2).PadLeft(set.Length, '0');
var cnt = Regex.Matches(Regex.Escape(result), "1").Count;
if (cnt == kElement) {
for(int j = 0; j < set.Length; j++)
if ( Char.GetNumericValue(result[j]) == 1)
Console.Write(set[j]);
Console.WriteLine();
}
}
}
为什么它能起作用?
在n元素集的子集和n位序列之间存在双射。
这意味着我们可以通过数数序列来计算出有多少个子集。
例如，下面的四个元素集可以用{0,1}X {0,1} X {0,1} X{0,1}(或2^4)个不同的序列表示。
我们要做的就是从1数到2^n来找到所有的组合。(我们忽略空集。)接下来，将数字转换为二进制表示。然后将集合中的元素替换为“on”位。
如果只需要k个元素的结果，则只在k位为“on”时打印。
(如果你想要所有的子集，而不是k长度的子集，删除cnt/kElement部分。)
(有关证明，请参阅麻省理工学院免费课件计算机科学数学，雷曼等，第11.2.2节。https://ocw.mit.edu/courses/electrical -工程-和-计算机- science/6 - 042 j -数学- -计算机科学-下降- 2010/readings/)
#include <stdio.h>
unsigned int next_combination(unsigned int *ar, size_t n, unsigned int k)
{
unsigned int finished = 0;
unsigned int changed = 0;
unsigned int i;
if (k > 0) {
for (i = k - 1; !finished && !changed; i--) {
if (ar[i] < (n - 1) - (k - 1) + i) {
/* Increment this element */
ar[i]++;
if (i < k - 1) {
/* Turn the elements after it into a linear sequence */
unsigned int j;
for (j = i + 1; j < k; j++) {
ar[j] = ar[j - 1] + 1;
}
}
changed = 1;
}
finished = i == 0;
}
if (!changed) {
/* Reset to first combination */
for (i = 0; i < k; i++) {
ar[i] = i;
}
}
}
return changed;
}
typedef void(*printfn)(const void *, FILE *);
void print_set(const unsigned int *ar, size_t len, const void **elements,
const char *brackets, printfn print, FILE *fptr)
{
unsigned int i;
fputc(brackets[0], fptr);
for (i = 0; i < len; i++) {
print(elements[ar[i]], fptr);
if (i < len - 1) {
fputs(", ", fptr);
}
}
fputc(brackets[1], fptr);
}
int main(void)
{
unsigned int numbers[] = { 0, 1, 2 };
char *elements[] = { "a", "b", "c", "d", "e" };
const unsigned int k = sizeof(numbers) / sizeof(unsigned int);
const unsigned int n = sizeof(elements) / sizeof(const char*);
do {
print_set(numbers, k, (void*)elements, "[]", (printfn)fputs, stdout);
putchar('\n');
} while (next_combination(numbers, n, k));
getchar();
return 0;
}
