leetcode-258-Add Digits

标签：leetcode

                                           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?

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.

将一个数赋值为该数的各位相加和，直到该数只有一位数时，输出。

暴力求解。

class Solution {
public:
    int addDigits(int num) {
        if (num < 10) return num;
        int tem;
        while (num > 9) {
            tem = num;
            num = 0;
            while (tem) {
                num += tem % 10;
                tem /= 10; 
            }
        }
        return num;
    }
};

找规律

例如，2,11,20,29  的最后答案都是2，他们与9取模都得2。

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

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

leetcode-258-Add Digits

标签：leetcode

原文地址：http://blog.csdn.net/u014705854/article/details/48132361