How to reverse a string in blocks of 2 in C++? - c++

Question

What I want to do is convert a string such as
"a4b2f0" into "f0b2a4"
or in more simple terms:
turning "12345678" into "78563412"
The string will always have an even number of characters so it will always divide by 2. I'm not really sure where to start.

Answer 1

One simple way to do that is this:
std::string input = "12345678";
std::string output = input;
std::reverse(output.begin(), output.end());
for(size_t i = 1 ; i < output.size(); i+=2)
std::swap(output[i-1], output[i]);
std::cout << output << std::endl;
Online demo
A bit better in terms of speed, as the previous one swaps elements twice, and this one swap each pair once:
std::string input = "12345678";
std::string output = input;
for(size_t i = 0, middle = output.size()/2, size = output.size(); i < middle ; i+=2 )
{
std::swap(output[i], output[size - i- 2]);
std::swap(output[i+1], output[size -i - 1]);
}
std::cout << output << std::endl;
Demo

Answer 2

Let's get esoteric... (not tested! :( And definitely not built to handle odd-length sequences.)
typedef <typename I>
struct backward_pair_iterator {
typedef I base_t;
base_t base;
bool parity;
backward_pair_iterator(base_t base, parity = false):
base(base), parity(parity) {
++base;
}
backward_pair_iterator operator++() {
backward_pair_iterator result(base, !parity);
if (parity) { result.base++; result.base++; }
else { result.base--; }
return result;
}
};
template <typename I>
backward_pair_iterator<I> make_bpi(I base) {
return backward_pair_iterator<I>(base);
}
std::string output(make_bpi(input.rbegin()), make_bpi(input.rend()));

Answer 3

static string reverse(string entry) {
if (entry.size() == 0) {
return "";
} else {
return entry.substr (entry.size() - 2, entry.size()) + reverse(entry.substr (0, entry.size() - 2));
}
}
My method uses the power of recursive programming

Answer 4

A simple solution is this:
string input = "12345678";
string output = "";
for(int i = input.length() - 1; i >= 0; i-2)
{
if(i -1 >= 0){
output += input[i -1];
output += input[i];
}
}
Note: You should check to see if the length of the string when mod 2 is = because otherwise this will go off the end. Do something like I did above.