Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.For example:Given n = 13,Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.Hint:1. Beware of overflow.解题思路：看得答案。很烧脑，没有看得很明白。这题实际上相当于一道找规律的题。那么为了找出规律，我们就先来列举下所有含1的数字，并每10个统计下个数，如下所示...
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.这道题是求一个数的数根。数根有一个同余性质：一个数与它的数根对(b-1)同余(b是进制数)。举个简单的例子就明白了：123=1*100+2*10+3=1*（99+1）+2*（9+1）+3=（99+2*9）+（1+2+3）前面一项能被9整除，后面...
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?题目意思是重复地将一个数字每一位数字加起来得到一个新的数字直到得到一个一位数。如果没有下面的提示O(1)，恐怕我也就直接模拟去...
Given an array nums of integers, return how many of them contain an even number of digits. Example 1:Input: nums = [12,345,2,6,7896]Output: 2Explanation: 12 contains 2 digits (even number of digits). 345 contains 3 digits (odd number of digits). 2 contains 1 digit (odd number of digits). 6 contains 1 digit (odd number of digits). 7896 contains 4 digits (even number of digits). Therefore only 12 an...
class Solution { public:int DigitSum(int num){int sum=0;do{sum=sum+num%10;num=num/10;}while(num!=0)return sum;}int addDigits(int num) {int n=num;while(n>=10){num=DigitSum(n); n=num;}return num;} };Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.哪里不对？原文：http://www.cnblogs.com/yuchenkit/p/5214670.html
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?给定一个非负整数，重复相加它的各个数位上的值直到这些数的和为1位数。按照题意使用while进行循环即可。class Solution { public...
258. Add DigitsDigit root 数根问题/*** @param {number} num* @return {number}*/var addDigits = function(num) {var b = (num-1) % 9 + 1 ;return b; };//之所以num要-1再+1;是因为特殊情况下：当num是9的倍数时，0+9的数字根和0的数字根不同。 性质说明 1.任何数加9的数字根还是它本身。（特殊情况num=0） 小学学加法的时候我们都明白，一个数字加9，就是把十位加1，各位减1。因此十位加个位的和是不变的；如果有进位，...
剑指 Offer 43. 1～n 整数中 1 出现的次数 233. Number of Digit Oneclass Solution { public:int countDigitOne(int n) {uint32_t base = 1;uint32_t ans = 0;int N = n;while(n) {ans += n / 10 * base;if(n % 10) ans += base;if(n % 10 == 1) ans -= base - N % base - 1;base *= 10;n /= 10;}return ans;} };
【LeetCode】738. Monotone Increasing Digits 单调递增的数字（Medium）（JAVA） 题目地址: https://leetcode-cn.com/problems/monotone-increasing-digits/ 题目描述： Given a non-negative integer N, find the largest number that is less than or equal to N with monotone increasing digits. (Recall that an integer has monotone increasing digits if and only if each pair of adjacent digits x and y satisfy x <= ...
这是悦乐书的第316次更新，第337篇原创01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第185题（顺位题号是788）。如果一个数字经过180度旋转后，变成了一个与原数字不同的数，这样的数被称为好数字。数字中的每一位都必须经过旋转。旋转的规则是：0,1,8这三个数旋转后还是自身，2旋转后变为5，5旋转后变为2,6旋转后变为9，9旋转后变为6，剩下的3，4，7旋转后并不能转成其他数字，是无效的。现在给出正数N，从1到N的好数字一...