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难度:容易
要求:
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
样例
Given num = 38.
The process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return 2.
思路:
public class Solution {
/**
* @param num a non-negative integer
* @return one digit
*/
public int addDigits(int num) {
// Write your code here
while(num > 9){
int tmp = 0;
while(num % 10 > 0 || num == 10){
tmp += num % 10;
num /= 10;
}
num = tmp;
}
return num;
}
}
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