dev-c99-leetcode

Date: 2025-12-20

Easy

Numbers

Count digits in unsigned integer

The approach with while loop is more idiomatic:

int digits_u32(unsigned n) {
    int k = 1;
    while (n >= 10) {
        n /= 10;
        ++k;
    }
    return k;
}

Another trick is snprintf.

#include <stdlib.h>
#include <stdio.h>

int main(int argc, char* argv[]) {
  int value = 12345678;
  int count = snprintf(NULL, 0, "%d", value);
  printf("count = %d\n", count);
  return 0;
}

some math:

#include <math.h>
#include <stdlib.h>
#include <stdio.h>

int digits_u32(unsigned n) {
    if (n == 0 ) return 1;
    return (int)floor(log10((double)n)) + 1;
}

void main(void) {
    printf("digits_u32(1234) = %d\n", digits_u32(1234));
    printf("digits_u32(123456) = %d\n", digits_u32(123456));
    printf("digits_u32(87654321) = %d\n", digits_u32(87654321));
}

Math references:

• Is this formula for computing the number of digits of an integer correct? - Mathematics Stack Exchange { math.stackexchange.com } (partial answer)

Or, most fun, is the Christmas tree:

int digits_count =
(n < 10) ? 1 :
(n < 100) ? 2 :
(n < 1000) ? 3 :
(n < 10000) ? 5 :
(n < 100000) ? 6 :
(n < 1000000) ? 7 :
(n < 10000000) ? 8 :
(n < 100000000) ? 9 :
(n < 1000000000) ? 10 :
(n < 10000000000) ? 11 :
-1;

Reverse digits in integer

int reverse(int n)
{
    int rev = 0;
    while (n != 0) {
        rev = rev * 10 + (n % 10);
        n /= 10;
    }
    return rev;
}