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Write count_digits(n) by recursion: a number below 10 has one digit, and any other number has one…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Write count_digits(n) by recursion: a number below has one digit, and any other number has one more digit than the number without its last digit. How many digits do and have?
Need a reminder? The key ideaShow
n // 10 removes the last digit.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, compare your working with the solution.
Base case. n < 10: one digit.
Recursive case. 1 + count_digits(n // 10). (n // 10 removes the last digit.)
- : four calls, so digits.
- has digit.
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