一看不让用循环就有点蒙。
只能找规律。和9相关。
class Solution {
public:
int addDigits(int num) {
if(num==0) return 0;
if(num%9==0) return 9;
return num%9;
}
}
};
原创
2022-08-05 15:58:46
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Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. Example: Follow up:Could you do it without any lo
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2019-10-28 09:19:00
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Given a non negative integer num, repeatedly add all its digits until the result has only one digit. Example: Input: 38 Output: 2 Explanation: The pro
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2018-10-05 11:11:00
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Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.For exam
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2022-08-03 16:56:37
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题目：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
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2023-03-07 12:33:50
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Given a non-negative integer num, repeatedly add all its digits until the result has only
one digit.
+ 8 = 11, 1 + 1 = 2. Since 2 has
on
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2023-09-05 09:28:23
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题目
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
Example:
Input: 38
Output: 2
Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2.
原创
2024-06-17 09:10:18
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int addDigits(int num){ while(num>=10){ int temp=0; while(num){ temp += num%10; num/=10; } num=temp; } return num; }
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2020-12-02 15:16:00
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【题目描述】给定一个非负整数 num，反复将各个位上的数字相加，直到结果为一位数。返回这个结果。https://leetcode.cn/problems/add-digits/description/【示例】【代码】Storm思路: num % 10 --> 求个位数  
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2023-01-07 20:36:39
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https://leetcode-cn.com/problems/add-digits/class Solution { public int addDig
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2022-08-19 15:38:57
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Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.For example:Given num = 38, the 2 has only one digit, r
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2023-06-07 15:50:09
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public int addDigits(int num) {if (num < 10) {return num;}int next = 0;while (num != 0) {next = next + num % 10;num /= 10;}return addDigits(next);}上边的递归很简单可以直接写成迭代的形式。public int addDigits(int num) {while (num >= 10) {int next = 0;while
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2023-05-15 16:43:50
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Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. Example: Input: 38 Output: 2 Explanation: The pro
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2019-10-17 23:58:00
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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:
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2016-08-31 21:48:00
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给定一个非负整数 num，反复将各个位上的数字相加，直到结果为一位数。 示例: 输入: 38输出: 2 解释: 各位相加的过程为：3 + 8 = 11, 1 + 1 = 2。 由于 2 是一位数，所以返回 2。进阶:你可以不使用循环或者递归，且在 O(1) 时间复杂度内解决这个问题吗？ class
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2020-09-29 19:01:00
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"题目" 同余定理，任何一个10进制数n 都可以表示成 n = a 10^x + b 10^(x 1) + .... c 10^0 那么 n ≡ ( a 10^x + b 10^(x 1) + .... c 10) mod 9 ( a 10^x + b 10^(x 1) + .... c 10) m
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2022-10-18 13:59:12
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problem 258. Add Digits solution1: solution2: Digital_root 参考 1. Leetcode_258. Add Digits; 2. Digital_root; 完
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2022-07-10 00:28:58
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DescriptionGiven a non-negative integer num, repeate
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2022-08-11 21:54:49
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给定一个非负整数 num，反复将各个位上的数字相加，直到结果为一位数。
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2022-12-07 14:11:08
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算法标签题目简叙思路代码class Solution {public: int addDigits(int nu
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2023-03-20 14:59:13
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