我试图在其中一个在线编码网站上解决 CountAndSay 问题，但我无法理解为什么我的程序正在打印NULL。我确信我在做一些概念上的错误，但没有得到它。
这是我的代码：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* countAndSay(int A) {
int i,j,k,f,count;
char a;
char *c = (char*)malloc(sizeof(char)*100);
char *temp = (char*)malloc(sizeof(char)*100);
c[0] = 1;c[1] = '\0';
for(k=2; k<=A; k++)
{
for(i=0, j=0; i<strlen(c); i++)
{
a = c[i];
count = 1;
i++;
while(c[i] != '\0')
{
if(c[i]==a)
{
count++;
i++;
}
else if(c[i] != a)
{
i--;
break;
}
else
{
break;
}
}
temp[j] = count;
temp[j+1] = a;
j += 2;
}
*(temp+j) = '\0';
if(k<A)
{
for(j=0; j<strlen(temp); j++)
{
c[j] = temp[j];
}
c[j] = '\0';
}
}
return temp;
}
int main(void) {
// your code goes here
char *c = countAndSay(8);
printf("%s\n",c);
return 0;
}
答案 0 :(得分：0)
这个想法并没有那么糟糕，主要的错误是数字和字符的混淆，如评论中所示。
另外：如果使用动态内存，则使用动态内存。如果您只想使用固定的小金额，则应使用堆栈，例如：c[100]，但评论中也会出现这种情况。您还只需要一块内存。以下是基于您的代码的工作示例：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// ALL CHECKS OMMITTED!
char *countAndSay(int A)
{
int k, count, j;
// "i" gets compared against the output of
// strlen() which is of type size_t
size_t i;
char a;
// Seed needs two bytes of memory
char *c = malloc(2);
// Another pointer, pointing to the same memory later.
// Set to NULL to avoid an extra malloc()
char *temp = NULL;
// a temporary pointer needed for realloc()-ing
char *cp;
// fill c with seed
c[0] = '1';
c[1] = '\0';
if (A == 1) {
return c;
}
// assuming 1-based input, that is: the first
// entry of the sequence is numbered 1 (one)
for (k = 2; k <= A; k++) {
// Memory needed is twice the size of
// the former entry at most.
// (Averages to Conway's constant but that
// number is not usable here, it is only a limit)
cp = realloc(temp, strlen(c) * 2 + 1);
temp = cp;
for (i = 0, j = 0; i < strlen(c); i++) {
//printf("A i = %zu, j = %zu\n",i,j);
a = c[i];
count = 1;
i++;
while (c[i] != '\0') {
if (c[i] == a) {
count++;
i++;
} else {
i--;
break;
}
}
temp[j++] = count + '0';
temp[j++] = a;
//printf("B i = %zu, j = %zu\n",i,j-1)
//printf("B i = %zu, j = %zu\n",i,j);
}
temp[j] = '\0';
if (k < A) {
// Just point "c" to the new sequence in "temp".
// Why does this work and temp doesn't overwrite c later?
// Or does it *not* always work and fails at one point?
// A mystery! Try to find it out! Some hints in the code.
c = temp;
temp = NULL;
}
// intermediate results:
//printf("%s\n\n",c);
}
return temp;
}
int main(int argc, char **argv)
{
// your code goes here
char *c = countAndSay(atoi(argv[1]));
printf("%s\n", c);
free(c);
return 0;
}
为了找到一种方法来检查不在OEIS列表中的序列，我在我的阁楼里翻找了一下这个小小的＆＃34;宝石＆＃34;：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
#include <limits.h>
char *conway(char *s)
{
char *seq;
char c;
size_t len, count, i = 0;
len = strlen(s);
/*
* Worst case is twice as large as the input, e.g.:
* 1 -> 11
* 21 -> 1211
*/
seq = malloc(len * 2 + 1);
if (seq == NULL) {
return NULL;
}
while (len) {
// counter for occurrences of ...
count = 0;
// ... this character
c = s[0];
// as long as the string "s"
while (*s != '\0' && *s == c) {
// move pointer to next character
s++;
// increment counter
count++;
// decrement the length of the string
len--;
}
// to keep it simple, fail if c > 9
// but that cannot happen with a seed of 1
// which is used here.
// For other seeds it might be necessary to
// use a map with the higher digits as characters.
// If it is not possible to fit it into a
// character, the approach with a C-string is
// obviously not reasonable anymore.
if (count > 9) {
free(seq);
return NULL;
}
// append counter as a character
seq[i++] = (char) (count + '0');
// append character "c" from above
seq[i++] = c;
}
// return a proper C-string
seq[i] = '\0';
return seq;
}
int main(int argc, char **argv)
{
long i, n;
char *seq0, *seq1;
if (argc != 2) {
fprintf(stderr, "Usage: %s n>0\n", argv[0]);
exit(EXIT_FAILURE);
}
// reset errno, just in case
errno = 0;
// get amount from commandline
n = strtol(argv[1], NULL, 0);
if ((errno == ERANGE && (n == LONG_MAX || n == LONG_MIN))
|| (errno != 0 && n == 0)) {
fprintf(stderr, "strtol failed: %s\n", strerror(errno));
exit(EXIT_FAILURE);
}
if (n <= 0) {
fprintf(stderr, "Usage: %s n>0\n", argv[0]);
exit(EXIT_FAILURE);
}
// allocate space for seed value "1" plus '\0'
// If the seed is changed the limit in the conway() function
// above might need a change.
seq0 = malloc(2);
if (seq0 == NULL) {
fprintf(stderr, "malloc() failed to allocate a measly 2 bytes!?\n");
exit(EXIT_FAILURE);
}
// put the initial value into the freshly allocated memory
strcpy(seq0, "1");
// print it, nicely formatted
/*
* putc('1', stdout);
* if (n == 1) {
* putc('\n', stdout);
* free(seq0);
* exit(EXIT_SUCCESS);
* } else {
* printf(", ");
* }
*/
if (n == 1) {
puts("1");
free(seq0);
exit(EXIT_SUCCESS);
}
// adjust count
n--;
for (i = 0; i < n; i++) {
// compute conway sequence as a recursion
seq1 = conway(seq0);
if (seq1 == NULL) {
fprintf(stderr, "conway() failed, probably because malloc() failed\n");
exit(EXIT_FAILURE);
}
// make room
free(seq0);
seq0 = NULL;
// print sequence, comma separated
// printf("%s%s", seq1, (i < n - 1) ? "," : "\n");
// or print sequence and length of sequence, line separated
// printf("%zu: %s%s", strlen(seq1), seq1, (i < n-1) ? "\n\n" : "\n");
// print the endresult only
if (i == n - 1) {
printf("%s\n", seq1);
}
// reuse seq0
seq0 = seq1;
// not necessary but deemed good style by some
// although frowned upon by others
seq1 = NULL;
}
// free the last memory
free(seq0);
exit(EXIT_SUCCESS);
}