I went ahead to see if I could complete the function on my own, thanks from some basic Java coding from before I actually completed it.
It will be interesting to compare it to what Michael will do, and I’m pretty sure he has a better way to solving this…
Codes can be solved in many ways, but the hard part is to do it as simple as possible (less code).
bool UBullCowCartridge::IsIsogram(FString Word)
{
for (size_t i = 0; i < Word.Len(); i++)
{
TCHAR x = Word[i];
for (size_t ii = i + 1; ii < Word.Len(); ii++)
{
TCHAR y = Word[ii];
if (y == x)
{
return false;
}
}
}
return true;
}
A run through to see if it behaves well.
