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leetcode 258_adddigits

admin 11月 12, 2020 0

Description

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.

Solution

有一种解法是利用循环完成：

class Solution {
public:
    int (int num) {
        while(num/10 >= 1){
            int sum = 0;
            int n = num;
            while(n > 0){
                sum += n%10;
                n /= 10;
            }
            num = sum;
        }
        return num;
    }
};

另一种解法，利用mod，小学的时候学过判断一个数能不能被9整除，直接将各位的数相加判断能不能被9整除，这个与本题有异曲同工之处，所以可以利用mod9的余数来表示。因为各个数相加不可能等于0，所以可以如下：

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