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[Leetcode]258. Add Digits(C++)

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:

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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