[LeetCode]: 258: Add Digits

题目：

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?

 

分析：

需要时间复杂度是O(1)，所以需要分析规律

经过计算1~30，得出结论：结果在1~9之间，提交代码为：

    public static int addDigits(int num) {
        int intRsult  = num%9;
        if(intRsult == 0){
            intRsult =9;
        }
        return intRsult;
    }

但是结果报错，发现遗忘了输入为0的情况，修改代码为：

public:  
    int addDigits(int num) {  
        return 1 + (num-1)%9;  
    }  
}; 

 

转载于:https://www.cnblogs.com/savageclc26/p/4779575.html