Problem
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
Example:
Input: 38
Output: 2
Explanation: 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?
Solution
/**
* @param {number} num
* @return {number}
*/
var addDigits = function(num) {
return 1 + (num - 1) % 9;
};
class Solution {
public int addDigits(int num) {
if (num == 0) {
return 0;
}
if (num % 9 == 0) {
return 9;
}
return num % 9;
}
}
Explain:
结果只有一个数字，就是 1~9 ，不断循环，找规律即可。
Complexity:
- Time complexity : O(1).
- Space complexity : O(1).