题目： 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?
思路： 其实就是如果这个数是0，返回0；这个数是9的倍数，直接返回9；若不是9的倍数，则返回它除9的余数。
代码：
class Solution {
public:
int addDigits(
int num) {
if(num==
0){
return 0;
}
else if(num%
9==
0){
return 9;
}
else{
return num%
9;
}
}
};
转载请注明原文地址: https://ju.6miu.com/read-11066.html
