# 258.Add Digits

**258.Add Digits**

难度:Easy

> 给定一个非负整数 num，反复将各个位上的数字相加，直到结果为一位数。

示例: 输入: 38 输出: 2 解释: 各位相加的过程为：3 + 8 = 11, 1 + 1 = 2。 由于 2 是一位数，所以返回 2。 进阶: 你可以不使用循环或者递归，且在 O(1) 时间复杂度内解决这个问题吗？

方法：最直接的就是循环相加，代码如下：

```
class Solution {
public:
    int addDigits(int num) {
        int res;
        while(num>9)
        {
            res=num%10;
            while(num>0)
            {
                num=num/10;
                res+=num%10;
            }
            num=res;
        }
        return res;
    }
};
```

方法2：使用归纳法列出数列的前面一些项，可以发现总是从1-9再循环，因此可以发现规律。

```
class Solution {
public:
    int addDigits(int num) {
        if(!num) return 0;
        int t=num%9;
        return t?t:9;
    }
};
```


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