我正在使用组合函数，我想知道是否有一种算法可以打印给定数组的所有子序列。也就是说，如果我给该算法指定序列“ ABCDEF”，它将打印：
A，
B，
C，
D，
E，
F，
AB，
AC，
广告，
AE，
自动对焦
公元前，
蓝光
是，
高炉
光盘，
CE，
CF，
DE，
DF，
EF，
美国广播公司
ABD，
阿伯
ABF，
ACD，
高手，
ACF，
ADE
ADF，
AEF，
BCD，
公元前
BCF，
BDE，
BDF，
BEF，
CDE，
CDF，
CEF，
DEF，
A B C D，
ABCE，
ABCF，
ABDE，
ABDF，
ABEF，
ACDE，
ACDF，
ACEF，
ADEF，
BCDE，
BCDF，
BCEF，
BDEF，
CDEF，
ABCDE，
ABCDF，
ABCEF，
ABDEF，
ACDEF，
BCDEF，
ABCDEF，
或者在更简单的情况下，如果我给它1234，它将打印：
1,2,3,4,12,13,14,23,24,34,123,124,134,234,1234。
如您所见，它不是一个任意排列，它只是子序列的最后一个成员的排列，其方式仍然是增加子序列。
我试图在c中创建一个函数来做到这一点，但是我真的很困惑，我的想法是使一个int L保持子序列的大小，而另一棵树将一个整数整数保持在子序列的头部，一个标志着与头部的分离，另一个滑动了给定数量的字符，但是却太混乱了。
有人可以帮我吗？
我的代码是：
int Stringsize( char m[] ){
int k;
for(k=0;;k++){
if( m[k] == '\0') break;
}
return (k-1);
}
void StringOrdM(char m[]){
int q,r,s,k;
for(k=0;k<=Stringsize(m);k++)
for(q=0;q<=Stringsize(m);q++)
for(s=q;s<=Stringsize(m);s++ )
printf("%c",m[q]);
for(r=q+1; r<=Stringsize(m) && (r-q+1)<= k ;r++ )
printf("%c", m[r] );
}
对于ABCD，它会打印A，A，A，A，B，B，B，C，C，D，AA，AB，AC，AD，BC，BD，CC，CD，DD等这是不对的，因为当它本来应该是A，B，C，D，AB，AC，AD，BC，BD，CD，...时，它会不断地重复A的4次，B的3次等等。 >
答案 0 :(得分：3)
正如我在上面的评论中所述，一种解决方案很简单：以二进制数计为(1<<n)-1。
因此，如果您有四个项目，则最多计数15个，其中 每个位模式都是元素的选择 。您将获得0001，0010，0011，0100，0101，0110，0111，{{1} }，1000，1001，1010，1011，1100，1101，1110。关于是否包含该数组元素，每一位都是一个真/假值。
1111如果您希望元素以最短到最长的顺序排序，则可以始终将这些结果存储在字符串数组中（使用#include <stdio.h>
int main(void) {
////////////////////////////////////////////////////////////////////////
int A[] = { 1, 2, 3, 4, 5 };
////////////////////////////////////////////////////////////////////////
size_t len = sizeof A / sizeof A[0]; // Array length (in elements)
size_t elbp = (1<<len) - 1; // Element selection bit pattern
size_t i, j; // Iterators
// Cycle through all the bit patterns
for (i = 1; i<=elbp; i++) {
// For each bit pattern, print out the 'checked' elements
for (j = 0; j < len; j++) {
if (i & (1<<j)) printf("%d ", A[j]);
}
printf("\n");
}
return 0;
}
），然后根据字符串长度进行排序（使用稳定的排序算法！）。
答案 1 :(得分：2)
我在上面的评论中提到，如果您不想使用位模式来查找所有排列，并根据想要的任何条件对结果进行排序，则还可以使用递归算法。
我怀疑这是一项家庭作业，您只要求一种算法，所以我留下了一些关键代码作为练习供您完成。但是，算法本身是完整的（关键部分仅在注释中描述，而不是插入功能代码）。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void printpermutations(const int *A, const size_t n, const char *pfix, const size_t rd);
int main(void) {
/////////////////////////////////////////////////////////////////////
int A[] = { 1, 2, 3, 4, 5 };
/////////////////////////////////////////////////////////////////////
size_t n = sizeof A / sizeof A[0]; // Array length (in elements)
size_t i; // Iterator
for (i = 1; i <= n; i++) {
printpermutations(A, n, "", i);
}
return 0;
}
// Recursive function to print permutations of a given length rd,
// using a prefix set in pfix.
// Arguments:
// int *A The integer array we're finding permutations in
// size_t n The size of the integer array
// char *pfix Computed output in higher levels of recursion,
// which will be prepended when we plunge to our
// intended recursive depth
// size_t rd Remaining depth to plunge in recursion
void printpermutations(const int *A, const size_t n, const char *pfix, const size_t rd) {
size_t i;
char newpfix[strlen(pfix)+22]; // 20 digits in 64-bit unsigned int
// plus a space, plus '\0'
if (n < rd) return; // Don't bother if we don't have enough
// elements to do a permutation
if (rd == 1) {
for (i = 0; i < n; i++) {
// YOUR CODE HERE
// Use printf() to print out:
// A string, consisting of the prefix we were passed
// Followed immediately by A[i] and a newline
}
}
else {
strcpy(newpfix, pfix);
for (i = 1; i <= n; i++) {
// YOUR CODE HERE
// Use sprintf() to put A[i-1] and a space into the new prefix string
// at an offset of strlen(pfix).
// Then, call printpermutations() starting with the ith offset into A[],
// with a size of n-i, using the new prefix, with a remaining
// recursion depth one less than the one we were called with
}
}
}
答案 2 :(得分：1)
根据torstenvl的回答，我编写了这段代码，它可以完美地工作。
如果有任何问题，请告诉我。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char str[] = "1234";
size_t len = strlen(str); // Array length (in elements)
char *A = malloc(sizeof(char) * len);
strcpy(A,str);
size_t elbp = (1<<len) - 1; // Element selection bit pattern
size_t i, j; // Iterators
int a = 0, b = 0, n = 0;
char **arr = malloc(sizeof(char*) * (10000)); //allocating memory
if (A[0] >= 'A' && A[0] <= 'Z') //If the string given is "ABCD...." transfer 'A' to '1' ; 'C' to '3' ...etc
for(int i = 0; i < len; i++)
A[i] = A[i] - 'A' + '1';
// Cycle through all the bit patterns
for (i = 1; i<=elbp; i++)
{
arr[b] = malloc(sizeof(char) * len);
// For each bit pattern, store in arr[b] the 'checked' elements
for (j = 0, a = 0; j < len; j++)
if (i & (1<<j))
arr[b][a++] = A[j];
b++;
}
int *num = calloc(sizeof(int) ,10000);
for (i = 0; i < b; i++)
num[i] = strtol(arr[i], NULL, 10); //convert char to int
for (i = 0; i < b; i++) //sort array numbers from smallest to largest
for (a = 0; a < i; a++)
if (num[i] < num[a])
{
n = num[i];
num[i] = num[a];
num[a] = n;
}
char *result = calloc(sizeof(char),10000);
for (i = 0, a = 0; i<b; i++)
a += sprintf(&result[a], "%d,", num[i]); //convert int to char and store it in result[a]
result[a - 1] = '\0'; //remove the last ','
len = strlen(result);
if (str[0] >= 'A' && str[0] <= 'Z') //if the string given is "ABCD..." transfer '1' to 'A' ; '12' to 'AB' ; '13' to 'AC'.....etc
for (i = 0; i < len; i++)
if(result[i] != ',')
result[i] = 'A' + (result[i] - '1') ;
///test
printf("%s",result);
return 0;
}
“ 1234”的输出：
1,2,3,4,12,13,14,23,24,34,123,124,134,234,1234
“ 123456789”的输出：
1,2,3,4,5,6,7,8,9,12,13,14,15,16,17,18,19,23,24,25,26,27,28,29,34,35,36,37,38,39,45,46,47,48,49,56,57,58,59,67,68,69,78,79,89,123,124,125,126,127,128,129,134,135,136,137,138,139,145,146,147,148,149,156,157,158,159,167,168,169,178,179,189,234,235,236,237,238,239,245,246,247,248,249,256,257,258,259,267,268,269,278,279,289,345,346,347,348,349,356,357,358,359,367,368,369,378,379,389,456,457,458,459,467,468,469,478,479,489,567,568,569,578,579,589,678,679,689,789,1234,1235,1236,1237,1238,1239,1245,1246,1247,1248,1249,1256,1257,1258,1259,1267,1268,1269,1278,1279,1289,1345,1346,1347,1348,1349,1356,1357,1358,1359,1367,1368,1369,1378,1379,1389,1456,1457,1458,1459,1467,1468,1469,1478,1479,1489,1567,1568,1569,1578,1579,1589,1678,1679,1689,1789,2345,2346,2347,2348,2349,2356,2357,2358,2359,2367,2368,2369,2378,2379,2389,2456,2457,2458,2459,2467,2468,2469,2478,2479,2489,2567,2568,2569,2578,2579,2589,2678,2679,2689,2789,3456,3457,3458,3459,3467,3468,3469,3478,3479,3489,3567,3568,3569,3578,3579,3589,3678,3679,3689,3789,4567,4568,4569,4578,4579,4589,4678,4679,4689,4789,5678,5679,5689,5789,6789,12345,12346,12347,12348,12349,12356,12357,12358,12359,12367,12368,12369,12378,12379,12389,12456,12457,12458,12459,12467,12468,12469,12478,12479,12489,12567,12568,12569,12578,12579,12589,12678,12679,12689,12789,13456,13457,13458,13459,13467,13468,13469,13478,13479,13489,13567,13568,13569,13578,13579,13589,13678,13679,13689,13789,14567,14568,14569,14578,14579,14589,14678,14679,14689,14789,15678,15679,15689,15789,16789,23456,23457,23458,23459,23467,23468,23469,23478,23479,23489,23567,23568,23569,23578,23579,23589,23678,23679,23689,23789,24567,24568,24569,24578,24579,24589,24678,24679,24689,24789,25678,25679,25689,25789,26789,34567,34568,34569,34578,34579,34589,34678,34679,34689,34789,35678,35679,35689,35789,36789,45678,45679,45689,45789,46789,56789,123456,123457,123458,123459,123467,123468,123469,123478,123479,123489,123567,123568,123569,123578,123579,123589,123678,123679,123689,123789,124567,124568,124569,124578,124579,124589,124678,124679,124689,124789,125678,125679,125689,125789,126789,134567,134568,134569,134578,134579,134589,134678,134679,134689,134789,135678,135679,135689,135789,136789,145678,145679,145689,145789,146789,156789,234567,234568,234569,234578,234579,234589,234678,234679,234689,234789,235678,235679,235689,235789,236789,245678,245679,245689,245789,246789,256789,345678,345679,345689,345789,346789,356789,456789,1234567,1234568,1234569,1234578,1234579,1234589,1234678,1234679,1234689,1234789,1235678,1235679,1235689,1235789,1236789,1245678,1245679,1245689,1245789,1246789,1256789,1345678,1345679,1345689,1345789,1346789,1356789,1456789,2345678,2345679,2345689,2345789,2346789,2356789,2456789,3456789,12345678,12345679,12345689,12345789,12346789,12356789,12456789,13456789,23456789,123456789
“ ABCDEF”的输出：
A,B,C,D,E,F,AB,AC,AD,AE,AF,BC,BD,BE,BF,CD,CE,CF,DE,DF,EF,ABC,ABD,ABE,ABF,ACD,ACE,ACF,ADE,ADF,AEF,BCD,BCE,BCF,BDE,BDF,BEF,CDE,CDF,CEF,DEF,ABCD,ABCE,ABCF,ABDE,ABDF,ABEF,ACDE,ACDF,ACEF,ADEF,BCDE,BCDF,BCEF,BDEF,CDEF,ABCDE,ABCDF,ABCEF,ABDEF,ACDEF,BCDEF,ABCDEF
答案 3 :(得分：0)
组合或k组合是从大小为n的集合中选择的k个元素的无序集合。 来源：http://www.martinbroadhurst.com/combinations.html 这是代码：
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;
}