【ACM从零开始】LeetCode OJ-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?

Hint:

A naive implementation of the above process is trivial. Could you come up with other methods?

What are all the possible results?

How do they occur, periodically or randomly?

You may find this Wikipedia article useful.

题目大意：给出一个非负数，将它每位数相加直到变成一个个位数。比如给出“38”，有“3”和“8”两位，相加得到“11”，再把“11”的每位相加，得到2，2即时最终解。

另外题目要求是否可以不使用循环来完成这个操作。

解题思路：这个问题很简单，先求出给出数字的位数个数，再通过递归去反复累加，直到数字小于10。

AC代码：

class Solution {
public:
　　int addDigits(int num) {
　 　　if (num < 10)
　　 　　return num;
　　　 int n = 0;
　　 int sum = 0;
　　 n = (int)log10(num);
　　 while (n >= 0)
　　 {
　　 　　 sum += num / (int)pow(10, n);
　　　　 num %= (int)pow(10, n);
　　　　 n--;
　　　}
　　　return addDigits(sum);
　 }　　
};