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[LeetCode] 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?
给定一个非负整数,重复相加它的各个数位上的值直到这些数的和为1位数。按照题意使用while进行循环即可。
class Solution { public: int addDigits(int num) { int sum = 0; while (num > 9) { while (num) { sum += (num % 10); num = num / 10; } num = sum; sum = 0; } return num; } }; // 3 ms
[LeetCode] Add Digits
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