从输入的字符数组中找到所有可能的单词（排列）
阅读了几篇文章之后，我仍然对排列和递归函数感到难过。 我试图在不等长的2D数组中创建所有可能的3个字母排列，其中第一个字母来自集合{‘l’，’m’，’n’}，第二个字母来自集合{‘q ‘，’r’}和第三个字母来自集合{‘a’，’e’，’i’，’o’}。 我的代码经过了正确的排列模式，但没有打印出正确的输出。 例如，如果前8个排列应该是：
lqa lqe lqi lqo lra lre lri lro 我的代码打印出来：
lqa e i o ra e i o
关于问题是什么的任何想法？ 以下是我的代码的相关部分：
rec(character_pools,0,3); void rec(char** pool, int k, int j) { if(k==j) { printf("\n"); return; } int i,len=strlen(pool[k]); for (i=0;i<len;i++) { printf("%c",pool[k][i]); rec(pool,k+1,j); } }
这最终为我工作！ Thx @Tsukuyo提示。 如果我想在排列中找到字符串的第n个索引，是否需要单独提问？
void rec(char** pool, int k, int j, char* cur, int counter) { if(k==j) { cur[k]=0; printf("Recursive call #%d %s\n",counter,cur); return; } int i,len=strlen(pool[k]); for (i=0;i
调用堆栈：
rec(pool, 0, 3); -> 'l' rec(pool, 1, 3); -> 'q' rec(pool, 2, 3); -> 'a' rec(pool, 3, 3); -> '\n' -> 'e' rec(pool, 3, 3); -> '\n' -> 'i' rec(pool, 3, 3); -> '\n' -> ... -> ... -> ...
更新：
不像递归那样。 但.. ..工作:)。 希望这可以帮助。
假设最大长度为10。
#include #include #define ALL_DONE 1 #define NOT_YET 0 int rec(char (*pool)[10], int num, int start); int main(void) { int i, num; char pool[20][10]; scanf("%d", &num); for (i = 0; i < num; i++){ scanf("%s", pool[i]); } while ( !rec(pool, num, 0) ); // keepint calling until all permutations are printed return 0; } int rec(char (*pool)[10], int num, int start) { static int ndx[20] = {0}; // record the index of each string if (start == num){ printf("\n"); return ALL_DONE; } printf("%c", pool[start][ndx[start]]); if ( rec(pool, num, start+1) == ALL_DONE ){ ndx[start+1] = 0; ndx[start]++; if (ndx[start] == strlen(pool[start])){ return ALL_DONE; return NOT_YET; } return NOT_YET; }
说明：
rec(pool, 0, 0)[1st calling] -> 'l' rec(pool, 1, 0) | ndx[0..2] = 0, 0, 0 -> 'q' rec(pool, 2, 0) | ndx[0..2] = 0, 0, 0 -> 'a' rec(pool, 3, 0) | ndx[0..2] = 0, 0, 0 -> '\n' retrun ALL_DONE | ndx[0..2] = 0, 0, 0 -> return NOT_YET | ndx[0..2] = 0, 0, 1 -> return NOT_YET | ndx[0..2] = 0, 0, 1 -> return NOT_YET | ndx[0..2] = 0, 0, 1 | rec(pool, 0, 0)[2nd] -> 'l' rec(pool, 1, 0) | ndx[0..2] = 0, 0, 1 -> 'q' rec(pool, 2, 0) | ndx[0..2] = 0, 0, 1 -> 'e' rec(pool, 3, 0) | ndx[0..2] = 0, 0, 1 -> '\n' return ALL_DONE | ndx[0..2] = 0, 0, 1 -> return NOT_YET | ndx[0..2] = 0, 0, 2 -> return NOT_YET | ndx[0..2] = 0, 0, 2 -> return NOT_YET | ndx[0..2] = 0, 0, 2 | | ... | rec(pool, 0, 0)[4th] -> 'l' rec(pool, 1, 0) | ndx[0..2] = 0, 0, 3 -> 'q' rec(pool, 2, 0) | ndx[0..2] = 0, 0, 3 -> 'o' rec(pool, 3, 0) | ndx[0..2] = 0, 0, 3 -> '\n' return ALL_DONE | ndx[0..2] = 0, 0, 3 -> return ALL_DONE | ndx[0..2] = 0, 0, 4 -> return NOT_YET | ndx[0..2] = 0, 1, 0 -> return NOT_YET | ndx[0..2] = 0, 1, 0 | | ... | ... | rec(pool, 0, 0)[5th] -> 'l' rec(pool, 1, 0) | ndx[0..2] = 0, 1, 0 -> 'r' rec(pool, 2, 0) | ndx[0..2] = 0, 1, 0 -> 'a' rec(pool, 3, 0) | ndx[0..2] = 0, 1, 0 -> '\n' return ALL_DONE | ndx[0..2] = 0, 1, 0 -> return NOT_YET | ndx[0..2] = 0, 1, 1 -> return NOT_YET | ndx[0..2] = 0, 1, 1 -> return NOT_YET | ndx[0..2] = 0, 1, 1 | | ... | rec(pool, 0, 0)[8th] -> 'l' rec(pool, 1, 0) | ndx[0..2] = 0, 1, 3 -> 'r' rec(pool, 2, 0) | ndx[0..2] = 0, 1, 3 -> 'o' rec(pool, 3, 0) | ndx[0..2] = 0, 1, 3 -> '\n' return ALL_DONE | ndx[0..2] = 0, 1, 3 -> return ALL_DONE | ndx[0..2] = 0, 1, 4 -> return ALL_DONE | ndx[0..2] = 0, 2, 0 -> return ALL_DONE | ndx[0..2] = 1, 0, 0 | | FINISH
您创建一个char数组，其中包含您要使用char str[] = "ABC";的字符串char str[] = "ABC";
然后你得到字符串的长度int n = strlen(str); 最后你置换了。
您创建一个新函数，它将包含输入字符串，字符串的起始索引和字符串的结束索引。 检查起始索引（ int s ）是否等于结束索引（ int e ），如果是，那意味着你已经完成，如果没有你进入一个循环，你从开始到结束（e），交换值，递归，再次交换回溯。
C ++中的一个例子：
#include #include void swap(char *i, char *j) { char temp; temp = *i; *i = *j; *j = temp; } void permutate(char *str, int start, int end) { int i; if (start == end) printf("%s\n", str); else { for (i = start; i <= end; i++) { swap((str + start), (str + i)); permutate(str, start + 1, end); swap((str + start), (str + i)); //backtrack } } } int main() { char str[] = "ABC"; int n = strlen(str); permutate(str, 0, n - 1); return 0; }