題目：

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:

A naive implementation of the above process is trivial. Could you come up with other methods?

What are all the possible results?

How do they occur, periodically or randomly?

class Solution { public:int addDigits(int num) {int res = 0;bool doneFlag = false;while(1){res = 0;while(num > 0){res += num%10;num /= 10;}if(res/10 == 0) break;num = res;}return res;} };
class Solution { public:int addDigits(int num) {if(num == 0) return 0;if(num % 9 == 0) return 9;return num%9;} };

總結(jié)

以上是生活随笔為你收集整理的[leetcode][math] Add Digits的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問(wèn)題。

如果覺(jué)得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。