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LeetCode258:Add Digits

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…

[LeetCode258] Add Digits 非负整数各位相加

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 has only one digit, return it. Follow up:Could you do it without…

[Swift]LeetCode258. 各位相加 | Add Digits

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. Since 2 has only one digit, return it. Follow up:Could you do…

LeetCode258 各位相加

题目链接:https://leetcode-cn.com/problems/add-digits/ 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? 常规思路: int addDigits(int x) { ) return x; ; while(x){ s…

public class Solution { public int AddDigits(int num) { var str = num.ToString(); ; foreach (var c in str) { result += Convert.ToInt32(c.ToString()); } ) { result = AddDigits(result); } return result; } } https://leetcode.com/problems/add-digits/#/de…

LeetCode258——Add Digits

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. 实现: class Solution { public: int…

LeetCode172 Factorial Trailing Zeroes. LeetCode258 Add Digits. LeetCode268 Missing Number

数学题 172. Factorial Trailing Zeroes Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. (Easy) 分析:求n的阶乘中末位0的个数,也就是求n!中因数5的个数(2比5多),简单思路是遍历一遍,对于每个数,以此除以5求其因数5的个数,但会超时. 考虑到一个数n比他小…