Unmash a string in C++ using recursion

[Hughng]

Homework Statement

MUST USE RECURSION TO SOLVE THESE PARTS.

Part A: Have a user input a string. Then display this string smashed up as follows: display the first character in the string, then the last, then the second, then the second to last, then the third... So if the string is “abcdef”, it will display:
afbecd (input “abcdef”)
12345 --> 15243
123456 --> 162534

Part B: Now, unmash the above strings.
i.e 162534 -->123456

Homework Equations

The Attempt at a Solution

I got part A to work.

#include <iostream>
using namespace std;
void mash(string s);

int main()
{
    string sequence;
    count << "Enter a sequence: ";
    getline(cin, sequence);
    mash(sequence);
}

void mash(string s)
{
    int a = s.length();
  
    if (a == 0)
    {
    return;
    }
    if (a == 1)
    {
    count << s;
    return;
    }
  
    count << s[0];  
  
    if(a>1)
    {          
        count << s[a - 1];  
        s = s.substr(1,a-2);
        mash(s);
      
    }
}

but I have no clue how to approach part B. I guess I can try to print out the characters in the even position, say in the string 162534, thus I will get 123. Then I guess I can try to print out the odd position characters from the last one up to the first one, i.e, 456. Combining these two will get the original strings but I have no clue how to use recursion to solve part B.

 

[mfb]

You can use recursion in the sub-steps: one recursion for the even characters, one for the odd characters.
Or unmash the string in the code with a recursive algorithm (2 characters at a time?), then print the whole string.

 

[NascentOxygen]

Combining these two will get the original strings but I have no clue how to use recursion to solve part B.

I'm looking forward to seeing the code you come up with to solve this. We can't help with improvements to your algorithm until you have first put in a good attempt by yourself.

Recursive routines pack a lot of power into few lines, so it's worth investing time in getting them right.