Add Digits
題意：
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?
解法一(循環)，程式碼如下：
public int addDigits(int num) {
if (num < 10) {
return num;
}
while (num >= 10) {
int temp = num;
int result = 0;
while (temp > 0) {
result += temp % 10;
temp /= 10;
}
num = result;
}
return num;
}
Time Complexity：待分解
網友 書影 提供以下觀查法
方法II：觀察法
根據提示，由於結果只有一位數，因此其可能的數字為0 - 9
使用方法I的代碼循環輸出0 - 19的運行結果：
in out in out
0 0 10 1
1 1 11 2
2 2 12 3
3 3 13 4
4 4 14 5
5 5 15 6
6 6 16 7
7 7 17 8
8 8 18 9
9 9 19 1
可以發現輸出與輸入的關係為：
out = (in - 1) % 9 + 1
public int addDigits2(int num) {
if (num < 10) {
return num;
}
return (num % 9 != 0) ? num % 9 : 9;
}
Time Complexity：$$O(1)$$