Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.
For example:
persistence(39) == 3 // because 39 = 27, 27 = 14, 1*4=4
// and 4 has only one digit
persistence(999) == 4 // because 999 = 729, 729 = 126,
// 126 = 12, and finally 1*2 = 2
persistence(4) == 0 // because 4 is already a one-digit number
Good Solution 1:
class Persist {
public static int persistence(long n) {
long m = 1, r = n;
if (r / 10 == 0)
return 0;
for (r = n; r != 0; r /= 10)
m *= r % 10;
return persistence(m) + 1;
}
}
Good Solution 2:
class Persist {
public static int persistence(long n) {
int times = 0;
while (n >= 10) {
n = Long.toString(n).chars().reduce(1, (r, i) -> r * (i - '0'));
times++;
}
return times;
}
}
