题目描述
题目链接:258. Add Digits
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
例子
Input: 38 Output: 2 Explanation: 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?
解题思路
简单地通过循环相加各个位直到最终 num < 9
即可,代码如下:
class Solution {
public:
int addDigits(int num) {
int result = 0;
while (num > 9) {
int next_num = 0;
while (num > 0) {
next_num += (num % 10);
num /= 10;
}
num = next_num;
}
return num;
}
};
- 时间复杂度: O(n)
- 空间复杂度: O(1)
GitHub 代码同步地址: 258.AddDigits.cpp
其他题目: GitHub: Leetcode-C++-Solution 博客: Leetcode-Solutions