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258. Add Digits
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?
不用循环
xyz = 100x + 10y +z, xyz % 9 = 99*x + 9*y + (x + y + z) % 9 = (x + y + z) % 9
java(3ms):
1 public class Solution { 2 public int addDigits(int num) { 3 if (num == 0) 4 return 0 ; 5 if (num > 9) 6 num %= 9 ; 7 if (num == 0) 8 num = 9 ; 9 return num; 10 } 11 }
258. Add Digits
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