carloscn
commented
1 year ago Analysis
static int32_t alternate_digit_sum(int32_t n)
{
int32_t ret = 0;
if (0 == n) {
return ret;
}
int32_t d = n, i = 0;
while (d != 0) {
ret += ((((i ++) & 0x1) ? -1 : 1)) * (d % 10);
d /= 10;
}
finish:
return ret;
}
Description
You are given a positive integer n. Each digit of n has a sign according to the following rules:
The most significant digit is assigned a positive sign. Each other digit has an opposite sign to its adjacent digits. Return the sum of all digits with their corresponding sign.
Example 1:
Input: n = 521 Output: 4 Explanation: (+5) + (-2) + (+1) = 4.
Example 2:
Input: n = 111 Output: 1 Explanation: (+1) + (-1) + (+1) = 1.
Example 3:
Input: n = 886996 Output: 0 Explanation: (+8) + (-8) + (+6) + (-9) + (+9) + (-6) = 0.
Constraints:
1 <= n <= 109