【LeetCode】258. Add Digits (2 solutions)
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?
解法一、按照定义做,直到结果只剩1位数字。
class Solution { public: int addDigits(int num) { int n = num; while(n > 9) { int cur = 0; while(n) { cur += (n % 10); n /= 10; } n = cur; } return n; } };
解法二、套公式digital root
class Solution { public: int addDigits(int num) { return num - 9 * floor((num - 1) / 9); } };