Given a non-negative integer num
, repeatedly add all its digits until the
result has only one digit.
For example:
Given num = 38
, the process is like: 3
,
+ 8 = 111 + 1 = 2
. Since 2
has
only one digit, return it.
实现:
class Solution {
public:
int addDigits(int num) {
int result = num;
while (true) {
if (result < 10) {
return result;
}
num = result;
result = 0;
while (num) {
result = result + num % 10;
num = num / 10;
}
}
}
};