E

1519709532

tags: Math

方法1: 普通做法就是按照题意, double-while loop把数字加起来. 第一层循环是O(n), 然后第二层循环就少很多数位, overall O(n)

方法2: 找数学规律, 每过9个数字, 取mod就会开始重复, 所以给所有数字取mod 就可以间接找到答案. O(1)

```

/*

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?

*/

/*

Thoughts:

Keep adding until result < 10

Double-while loop: start with a num, calculate result, then num = result.

*/

class Solution {

public int addDigits(int num) {

if (num < 10) {

return num;

}

while (num > 9) {

int temp = 0;

while (num != 0) {

temp += num % 10;

num = num / 10;

}

num = temp;

}

return num;

}

}

class Solution {

public int addDigits(int num) {

return (num - 1) % 9 + 1;

}

}

```