Add Digits 解答

Question

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 = 111 + 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:

    1. A naive implementation of the above process is trivial. Could you come up with other methods?
    2. What are all the possible results?
    3. How do they occur, periodically or randomly?
    4. You may find this Wikipedia article useful.

Solution

Naive way is to use two loops.

 1 public class Solution {
 2     public int addDigits(int num) {
 3         int copy = num;
 4         while (copy / 10 > 0) {
 5             int newNum = 0;
 6             while (copy / 10 > 0) {
 7                 newNum += copy % 10;
 8                 copy /= 10;
 9             }
10             newNum += copy;
11             copy = newNum;
12         }
13         return copy;
14     }
15 }

There is a math relationship.

1 public class Solution {
2     public int addDigits(int num) {
3         if (num == 0)
4             return num;
5         if (num % 9 == 0)
6             return 9;
7         return (num % 9);
8     }
9 }
原文地址:https://www.cnblogs.com/ireneyanglan/p/4870599.html