Reputation: 199
I have to input a number n
, an a
digit and a b
digit and output the number n
with all the a
digits in it replaced by a b
one. For example:
Input:
n = 1561525
a = 5
b = 9
Output:
n = 1961929
Should be recursive ! I didn't post any code as I've done it in a non-recursive way but apparently it's not even close to what I need.
Thanks for the help !
Upvotes: 1
Views: 3392
Reputation: 2373
Check this, it works but maybe it is to much C
int convert(int num, int a, int b)
{
if( num )
{
int res = convert(num/10,a,b);
int t = num%10;
res *=10;
t = t == a ? b:t;
res = res + t;
return res;
}
return 0;
}
Divide by 10 the initial number, until nothing left of it, and then construct it again replacing a with b.
Upvotes: 1
Reputation: 39800
So the easiest and safest way I can think of, is by using std::replace
:
int replace(int num, int d1, int d2) {
string s = std::to_string(num); // convert to string
std::replace( s.begin(), s.end(), d1+'0', d2+'0'); // call std::replace
return atoi( s.c_str() ); // return the int
}
Now if you really have to use recursion (there is no need for it here), here's one possible solution:
using std::string;
// recursive function, accepts a string, current index, c2 replaces c1
string replace_rec (string s, unsigned index, char c1, char c2) {
// check if the it's still a valid index
if (index < s.size()) {
// if this is a char to be converted, do so
if (s[index] == c1)
s[index] = c2;
// call itself but with an updated string and incremented index
replace_rec(s, index+1, c1, c2);
}
// the last call will result in the string with all chars checked. return it
return s;
}
// call this function with input, the num to be replaced and what with
int replace(int num, int d1, int d2) {
string s = std::to_string(num); // convert to string
// convert the result back to int and return it.
return atoi( replace_rec(s, 0, d1+'0', d2+'0').c_str() );
}
In any case, you can call your replace()
function like this:
int main(){
cout << replace (4578, 4, 9); // output: 9578
cin.get();
}
Upvotes: 1
Reputation: 236004
To make things easier, you can convert the number into a string (a char[]
in C++). Then, it's a simple matter of iterating over it and checking at each step if the number we want to replace was found in the current position. For a possible solution, here's an implementation of the algorithm in Python - one of the nice things of the language is that it reads almost as pseudocode, and it should be relatively simple to port to C++:
def aux(n, a, b, i):
if i == len(n):
return ''
elif n[i] == a:
return b + aux(n, a, b, i+1)
else:
return n[i] + aux(n, a, b, i+1)
def change(n, a, b):
return int(aux(str(n), str(a), str(b), 0))
It works as expected:
change(1561525, 5, 9)
=> 1961929
Upvotes: 1