【算法题】字符串的全排列

void DoPermute(char in[],char out[],int used[],int length,int recursLev)
{
int i;
/*基底情况*/
if(recursLev==length)
{
printf("%s\n",out );
return ;

}

/*递归情况*/
for(i=0;i<length;i++)
{
if(used[i])
continue;
out[recursLev]=in[i];
used[i]=1;
DoPermute(in,out,used,length,recursLev+1);
used[i]=0;
}
}

int Permute(char inString[])  //字符串全排列
{
int length,i,*used;
char *out;
length=strlen(inString);
out=(char *)malloc(length+1);
if(!out)
return 0;
out[length]='\0';
used=(int *)malloc(sizeof(int)*length);
if(!used)
return 0;
/*初始化*/
for(i=0;i<length;i++)
{
used[i]=0;
}
DoPermute(inString,out,used,length,0);
free(out);
free(used);
return 1;
}
void main()
{
char *s="abcd";
Permute(s);

}