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?
将一个数的各个位加起来,如此反复知道只剩各位数。不用循环和递归在常数时间复杂度内解决。
列几个数就可以发现规律,除了0以外其他数以1-9循环。
代码语言:javascript复制class Solution {
public:
int addDigits(int num) {
return 1 (num - 1) % 9;
}
};