c - 如何获得返回数组而不是打印数组的函数？

Question

该函数写在下面。我希望此函数返回排列后的字符串的数组而不是打印它，因为我的目标是能够查看给定字符串可能有多少排列，并且要求使用递归。

void permuteString(char *a, int first, int last) //parameters are array, first index, and final index //requires a swap function
{
    int i;

    if (first == last) //this is the base case that indicates that there are no more letters to be permuted
    {
        printf("%s\n", a);
    }

    else
    {
        for(i = first; i <= last; i++)
        {
            swap((a + first), (a + i));
            permuteString(a, first + 1, last);
            swap((a + first), (a + i));
        }
    }
}

Answer 1

我了解您需要返回字符数组而不是打印字符数组，因此，请首先更改函数的返回类型并按如下所示返回字符数组，

        Char *permuteString(char *a, int first, int last) //parameters are array, first index, and final index //requires a swap function
        {
            int i;

            if (first == last) //this is the base case that indicates that there are no more letters to be permuted
            {
               // printf("%s\n", a);
                return a;
            }

            else
            {
                for(i = first; i <= last; i++)
                {
                    swap((a + first), (a + i));
                    permuteString(a, first + 1, last);
                    swap((a + first), (a + i));
                }
            }
        }