Add Digits

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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?

假设num的各位数是a,b,c

则num = a*100 + b*10 + c = a+b+c + a*99+b*9+0

那么num%9 =  (a+b+c + a*99+b*9+0)%9 = (a+b+c)%9

所以只要用num对9取模就是可能的结果。如果num%9=0说明a+b+c=9，那么返回9,否则返回num%9

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

Add Digits

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原文地址：http://www.cnblogs.com/zengzy/p/5002313.html