LeetCode oj 258. Add Digits(数字根）

258. Add Digits

 

Question
Editorial Solution
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Total Accepted: 127957
Total Submissions: 257201
Difficulty: Easy

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?

给你一个数，将这个数的各个位相加，直到这个数只有一位

经典的数字根问题，之前做过，就是看能不能除开9，下面给出具体证明

设一个数是abcde，则abcde = a * 10000 + b * 1000 + c * 100 + d * 10 + e,将这个式子拆一下得到(a + b + c + d + e) + (a * 9999 + b * 999 + c * 99 + d * 9),

显然右侧括号的式子可以整除9，所以num % 9的式子成立

需要注意的是特判0 和 9

public int addDigits(int num) {
if(num % 9 == 0){
if(num == 0)
return 0;
else
return 9;
}
while(num >= 9){
num %= 9;
}
return num;
}