https://leetcode.com/problems/number-of-digit-one/ Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.Example:Input: 13 Output: 6 Explanation: Digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.代码：class Solution { public:int countDigitOne(int n) {int res = 0;long long a = 1, b = 1;while (n) {res += (n + 8) / 10 * a + (...
The Fibonacci sequence is defined by the recurrence relation: Fn = Fn?1 + Fn?2, where F1 = 1 and F2 = 1.Hence the first 12 terms will be: F1 = 1 F2 = 1 F3 = 2 F4 = 3 F5 = 5 F6 = 8 F7 = 13 F8 = 21 F9 = 34 F10 = 55 F11 = 89 F12 = 144The 12th term, F12, is the first term to contain three digits.What is the index of the first term in the Fibonacci sequence to contain 1000 digits?找到第一个1000位的斐波...
Given a string ‘str’ of digits and an integer ‘n’, build the lowest possible number by removing ‘n’ digits from the string and not changing the order of input digits.Examples:Input: str = "4325043", n = 3 Output: "2043"Input: str = "765028321", n = 5 Output: "0221"Input: str = "121198", n = 2 Output: "1118"解法一：目标是要得到用字符串表示的最小数字，所以要使高位的数字尽量小，可以使用贪心策略...
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.public class Solution {public int countDigitOne(int n) {long res = 0;int left = n;int right = 0;int base = 1;while (left > 0) {int cur = left % 10;left = left / 10;right = n % bas...
题目如下：Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.Example:Input: 13 Output: 6 Explanation: Digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.解题思路：本题是《编程之美》上的题目，详细分析如下图。代码如下：class Solution(object):def countDigitOne(self, n):""":type n: int:rtype: int"""if n <= 0:return 0res = 0sn...
用hashMap查重：代码：public class Solution { public boolean isHappy(int n) { Map<Integer, Integer> map = new HashMap<>(); while(!map.containsKey(n)){ map.put(n,1); int result = 0; result += (n%10) * (n%10); while(n/10 > 0){ n /= 10; result += (n%10) * (n%10); } n = result; } ...
1295. Find Numbers with Even Number of Digits（统计位数为偶数的数字）链接https://leetcode-cn.com/problems/find-numbers-with-even-number-of-digits题目给你一个整数数组?nums，请你返回其中位数为?偶数?的数字的个数。示例 1：输入：nums = [12,345,2,6,7896] 输出：2 解释： 12 是 2 位数字（位数为偶数）? 345 是 3 位数字（位数为奇数）?? 2 是 1 位数字（位数为奇数）? 6 是 1 位数字 位数为奇数）? 7896 是 4 位数字（...
转自：http://blog.csdn.net/renfufei/article/details/41648061问题描述昨天,以及今天(2014-11-29),使用 TortoiseGit 时碰到了一个诡异的问题. 卸载，清理注册表,重装，重启，各种折腾以后,还是不能解决. 但是23.45分一过,突然灵光一闪,解决了.问题是这样的. 使用命令行的 git push, git fetch, git pull 什么的都没问题. 但是使用 TortoiseGit 执行拉取(pull ...) 命令时, 就给报错, 报错信息如下: [plain] view plaincopy git.e...
题目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.解题思路题目理解：题目的意思是给定一个数n，计算[0, n]的区间中1出现的次数。具体思路：在面试过程中如果被问到类似的题目，一定不要着急，因为这肯定是一道找规律的题目。最好的办法...
Number of Digit OneGiven 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. 数学题，真是为难了数学拙计的我了。递归分治，拿8192举栗子：把8192拆成：1-999 -> 递归(999)1000-1999 -> 1000个1 + 递归(999)2001-2999 -> 递归(999)..8000-8192 ->...
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...
问题描述昨天,以及今天(2014-11-29),使用 TortoiseGit 时碰到了一个诡异的问题. 卸载，清理注册表,重装，重启，各种折腾以后,还是不能解决. 但是23.45分一过,突然灵光一闪,解决了.问题是这样的. 使用命令行的 git push, git fetch, git pull 什么的都没问题. 但是使用 TortoiseGit 执行拉取(pull ...) 命令时, 就给报错, 报错信息如下: [plain] view plaincopy git.exe pull -v --no-rebase --progress "origin" /libexec/git-c...
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...
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;} };
正如标题所示,我需要一个C#equivelant of ROUNDDOWN. 例如,如果你采用图13.608000,我正在寻找的输出是13.60. 我似乎无法找到任何涵盖我所追求的东西.解决方法:您可以执行以下操作：var rounded = Math.Floor(13.608000 * 100) / 100;请注意,Math.Floor()向下舍入到最接近的整数,因此需要乘以,向下舍入,然后除以.