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?
Method:
Digit root is the number that add all digits until the result has only one digit.
C++
class Solution {
public:
int addDigits(int num) {
return (num-1)%9+1;
}
};
LeetCode 258. Add Digits Description Given a non-negative...
转自我的 blog: 用 Swift 刷 LeetCode No.258 - Add Digits 用 Swift...
Given a non-negative integer num, repeatedly add all its ...
原题 给出一个非负整数，重复计算为一位的和，直到这个和变为一位数 样例：给出 num = 38，过程即：3 + 8...
Given a non-negative integer num, repeatedly add all its ...
Add Digits 题目描述： Given a non-negative integer num, repeat...
D69 258. Add Digits 题目链接 258. Add Digits 题目分析 给定一个数字，给每一位...
原题链接：Add Digits 这是一道数学题，代码如下： 我不讲解这道题，大家直接看下面这张图就能明白了。
Given a non-negative integer num, repeatedly add all its ...
本文标题：[LeetCode 258] Add Digits
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