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?
class Solution {
func addDigits(num: Int) -> Int {
if num == 0 {
return 0
}
return (num - 1) % 9 + 1
}
}
Problem Description Given a non-negative integer num, rep...
Question: Given a non-negative integer num, repeatedly ad...
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。
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...
题目链接：7. Reverse Integer难度：Easy Reverse digits of an integ...
LeetCode 258. Add Digits Description Given a non-negative...
题目： Given a non-negative integer num, repeatedly add all ...
258. Add Digits[思路]数字累加，将给定的一个整数，将个位，十位，百位等相加，连续操作，直到最后的值...
循环： 这个办法很神奇
本文标题：2、Add Digits（Easy）
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