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 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
Follow up:
Could you do it without any loop/recursion in O(1) runtime?
class Solution {
public:
int addDigits(int num) {
int ans = num;
// while(ans!=2)
// {
while()
{
if(ans < ) return ans;
num = ans;
ans = ;
while(num != )
{
ans += num%;
num /= ;
}
}
}
};
O(1)方法
观察如下输入：
输入：1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
输出：1,2,3,4,5,6,7,8,9,1 , 2 , 3 , 4 , 5 , 6, 7 , 8 , 9 , 1 , 2
可以看到返回的值就是(num-1)%9+1