Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

Example:

Input: 38Output: 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?

题意：把2一个数拆开，最终变成一个1-9的数字

最先看到题先想到的是递归拆，勉强过了

Java

1  public int addDigits(int num) { 2         System.out.println(num); 3         if (num <= 9) 4             return num; 5         int sum = 0; 6         while (num != 0) { 7             sum += num%10; 8             num = num/10; 9         }10         return addDigits(sum);11     }12 }

显然题意不可能是用这个算法，末尾也有提示说O(1) 的算法复杂度，那就只能找规律了

1-9   10 11 12 13 18  19-27

1-9   1-9                       1-9

规律就比较明显了

Java

1 class Solution {2 3     public int addDigits(int num) {4         return (num-1)%9 + 1;5     }6 }