【LeetCode】258. Add Digits

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本文介绍了一种数字压缩算法，该算法通过不断累加给定非负整数的所有位数直到结果仅剩一位数。示例中，输入数字38，经过两轮累加得到最终结果2。文章挑战读者思考如何在O(1)时间内不使用循环或递归实现该算法。

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?

Show Hint

按照定义做，直到结果只剩1位数字。

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