LeetCode 258. Add Digits(1+(n-1)%9)
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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?
思路:ans(n) = 1 + (n - 1) % 9
Code:
class Solution {public: int addDigits(int num) { return 1 + (num - 1) % 9; }};
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