258. Add Digits

题目：

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?

Hint:

1. A naive implementation of the above process is trivial. Could you come up with other methods?
2. What are all the possible results?
3. How do they occur, periodically or randomly?
4. You may find this Wikipedia article useful.

链接： http://leetcode.com/problems/add-digits/

题解：

又是数学题，求digital root。循环叠加比较容易，但看了wiki以后发现了公式，还是用公式算吧。这种数学题对数学不好的我来说真是头大。原理10 % 9 或者 100 % 9都等于 1 % 9。举个例子n = abc = a  * 100 + b * 10 + c，那么 (a*100 + b * 10 + c) % 9 = (a + b + c) % 9。由此n == 0时，result = 0， n % 9 == 0时， 说明a + b + c = 9，我们返回9，对于其他数字， (a + b + c)等于res % 9。  

Time Complexity - O(1)， Space Complexity - O(1)

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

 

二刷：

Java:

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

 

三刷:

发现前两刷其实并没有完全理解，也许就是看了discuss区的答案而已。为什么(a + b + c) mod 9 = (abc) mod 9, 真正用到的公式是modulo运算的分配和结合律。

1.  (a + b) mod n  = ((a mod n) + (b mod n)) mod n

2.  (a * b) mod n = ((a mod n) * (b mod n)) mod n

假如一个数字的三位字符是abc，那么这个数等于 a * 100 + b * 10 + c, 根据分配律，  (a * 100) mod 9 = ((a mod 9) * (100 mod 9)) mod 9 = a mod 9，b和c同理， 所以 (a * 100 + b * 10 + c) mod 9 = (a + b + c) mod 9。  我们还可以使用一个小技巧，再用一次分配律直接用 (num - 1) mod 9 + 1来得到结果，这样可以避免一些边界条件的判断。

 

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

 

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

 

Update:

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

 

 

Reference:

https://en.wikipedia.org/wiki/Digital_root

https://en.wikipedia.org/wiki/Modulo_operation

https://leetcode.com/discuss/67755/3-methods-for-python-with-explains

https://leetcode.com/discuss/52122/accepted-time-space-line-solution-with-detail-explanations

https://leetcode.com/discuss/55910/two-lines-c-code-with-explanation