leetcode 258 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?

 

 

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

設數字爲四位數，各個位分別是abcd，其他情況類似。

 

則

abcd=a*1000+b*100+c*10+d

        =a+b+c+d+(a*999+b*99+c*9)

        =a+b+c+d+(a*111+b*11+c)*9

設n=a+b+c+d

n同樣可以做如上展開，反覆迭代，直到前面數字之和爲個位數，形如n+m*9並且此時這個個位數的取得滿足題目的要求的返回值，此時後面的數字爲9的倍數。要求的這個個位數只需對九求模即可（因爲這個數可能是9，所以要先減一最後加一）。