Add Digits

Add Digits

Difficulty: Easy

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?

Hint:

A naive implementation of the above process is trivial. Could you come up with other methods?

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最直接的思路：

num2用来统计num有多少位；

num1用来做除数。

eg.num为345，取百位，则num1得为1000（取余后，/（num1/10）即得百位数3）

int addDigits(int num) {
int num1,num2;
num1=1,num2=0;
int sum=0;
if(num/10 == 0)
return num;
while(num1 <= num){
num1 *= 10;
num2++;
}
while(num2--){
num %= num1;
num1 /= 10;
sum += num/num1;
}
addDigits(sum);
}

最终结果显示：Last executed input:

2032610959

Status: Time Limit Exceeded

显然该法虽直接，但运算效率太低。

于是我们对这些数做了下计算：

0-9，全部都各自返回本身（0-9）

10 -> 1

11 - > 2

……

19 - > 2

20 - > 2

21 - >　３

……

发现规律，最终结果的区间总落在[0，9]之间。

21%9余3，20%9余2，19%9余1，18%9余0，但实际上18对应结果为9.

所以应该总结为(num-1)%9+1;

int addDigits(int num) {
return (num-1)%9+1;
}