Function to manipulate a string ("abcdef" -> "faebdc") - c++

Question

Hi everyone I'm working on a function to manipulating any string in this following manner.
"abc" -> "cab"
"abcd" -> "dacb"
"abcdef" -> "faebdc"
"divergenta" -> "adtinveerg"
... and so on.
This is the code I've come up with so far. I think it does the job but I think the code and solution is kind of ugly and I'm not sure if it's fail proof and if it is working properly for every given case. I would highly appreciate any input on this code or any examples on how you would write this function. I beg you to bear in mind that I'm very much a n00b so don't go too hard on me.
string transformer(string input) {
string temp;
int n = 0;
int m = (input.length() -1);
for( int i = 0; i < input.length(); i++) {
temp += input[m];
if (input[m] == input[n]) {
break;
}
else {
temp += input[n];
}
n += 1;
m -= 1;
if ( temp.length() == input.length() ) {
break;
}
}
return temp; }

Answer 1

You have three problems.
Try it with "abbba". If the result isn't what you want, then this conditional:
if (input[m] == input[n]) {
break;
}
is just plain wrong.
Look at the other conditional:
if ( temp.length() == input.length() ) {
break;
}
You're adding two characters at a time to temp. What if input has odd length?
Suppose that works correctly. Consider the loop:
for( int i = 0; i < input.length(); i++) {
...
if ( temp.length() == input.length() ) {
break;
}
}
That loop will never terminate in the for statement. You might as well do it this way:
while( temp.length() < input.length() ) {
...
}
Once that's all working correctly, you should look into iterators.

Answer 2

This function just walks two indices toward the center until they meet or pass each other. The last if block handles the case of an odd length input string. It works for all your test cases on ideone.com
std::string transformer(const std::string& input)
{
std::string temp;
int i = 0;
int j = input.length() - 1;
while (i < j) {
temp += input[j--];
temp += input[i++];
}
if (i == j) {
temp += input[i];
}
return temp;
}

Answer 3

std::string transformer(const std::string& input) {
std::string res(input.length(), '0');
for (int i = 0; i < input.length(); ++i) {
res[i] = input[ i % 2 == 0 ? input.length() - (i/2) - 1 : (i/2) ];
}
return res;
}

Answer 4

Unfortunately if (input[m] == input[n]) will make sure that if the first and last characters are the same, it immediately quits after the first character is processed.

Answer 5

I'd do this with std::string::iterator and std::string::reverse_iterator:
auto it = input.begin();
auto rit = input.rbegin();
std::string temp;
for (size_t i = 0; i < input.length()/2; ++i) {
temp += *rit++;
temp += *it++;
}
The logic for handling empty and odd-length input is left for you to do, shouldn't be too hard. (Input of length 1 is also a special case)

Answer 6

I would use pointers instead of indexes to do this.
So you have a pointer the reading the edges and you keep swapping them with each iteration.
It will also make it faster.
I think this should work, but I can't remember how to make an array of const char pointers. Can anyone help me with that step?
string transformer(string input) {
std::string temp;
const char *front, *back;
for (*front = input.c_str(), *back = front + input.length() - 1; front < back ; front++, back--) {
temp += *back;
temp += *front;
}
if (front == back)
temp += *front;
return temp;
}
(using the same method as #Blastfurnace, but skipping unnecessary indexes.)