1.算法说明: 如3141592,在m(digitDivide)=100时,即要求计算百位上"1"的个数 其中a为31415,b为92,31415中出现了3142次"1",因为每10个数里面出现1次"1".而实际上,31415是3141500,所以把31415中1的个数再乘以m.如3141400~3141499中,前缀为31414的数出现了100次,所以需要乘以m(此时是100). class Solution { public: int cou…

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. 解法:参考编程之美 132页 2.4 1的数目,以下代…

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: Beware of overflow.…

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: Beware of overflow. 大…

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: Beware of overflow. i…

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: 递归 static public in…

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: Beware of overflow. c…

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. 链接: http://leetcode.com…

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. 代码: class Solution {…

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. 给定一个整数 n,计算所有小于等于 n 的非负整数中数字…

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. 按不同位置统计 31456 统计百位时: (0-31)…

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. 方法一 class Solution { public:…

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. 思路: 对这个数字的每一位求存在1的数字的…

原题链接在这里:https://leetcode.com/problems/number-of-digit-one/ 每10个数, 有一个个位是1, 每100个数, 有10个十位是1, 每1000个数, 有100个百位是1.  做一个循环, 每次计算单个位上1得总个数(个位,十位, 百位). 例子: 以算百位上1为例子:   假设百位上是0, 1, 和 >=2 三种情况: case 1: n=3141092, a= 31410, b=92. 计算百位上1的个数应该为 3141 *100 次. c…

这题属于需要找规律的题.先想一下最简单的情形:N = 10^n - 1 记X[i]表示从1到10^i - 1中 1 的个数,则有如下递推公式:X[i] = 10 * X[i - 1] + 10^(i - 1) 这个递推公式可以这么观察得到: i = 0, X[0] = 0 i = 1, 从1到9, X[1] = 1 i = 2, 从1到99, X[2] = 20:可以设想,把所有数都写成两位数(比如1写成01, 2写成02),我们暂且不统计最高位的1, 则首先至少有10 * X[1]个1,然后我…

题目意思:一个int数组,有一个数只出现一次,其他数均出现两次,找到这个唯一数 知识普及:~:非运算,单目运算符1为0,0为1;   &:与运算,都为1则为1,否则为0 |:或运算,全为0则为0,否则为1        ^:异或运算,相同为0,不同为1 思路:将数组中元素进行异或运算,则只剩下0和唯一数,异或得到的是唯一数 class Solution { public: int singleNumber(vector<int>& nums) { ; ;i<nums.si…

给定一个整数 n,计算所有小于等于 n 的非负数中数字1出现的个数. 例如: 给定 n = 13, 返回 6,因为数字1出现在下数中出现:1,10,11,12,13. 详见:https://leetcode.com/problems/number-of-digit-one/description/ Java实现: 方法一: class Solution { public int countDigitOne(int n) { StringBuilder sb=new StringBuilder()…

All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017) .   Top Interview Questions # Title Difficulty Acceptance 1 Two Sum Medium 17.70% 2 Add Two N…

hihoCoder #1432 : JiLi Number(吉利数) 时间限制:1000ms 单点时限:1000ms 内存限制:256MB Description - 题目描述 Driver Ji likes the digit "1". He has an accumulator which shows the sum of input number. He lists all of positive number no more than N and starts counting…

题目连接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=247&page=show_problem&problem=3666 13764622 1225 Digit Counting Accepted C++11 0.035 2014-06-18 07:44:02 1225 - Digit Counting Time limit: 3.000 seconds Tru…

Write an algorithm to determine if a number is "happy". A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the…

Single Number II Given an array of integers, every element appears threetimes except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 解法一:开辟map记录次数 class So…

Trung is bored with his mathematics homeworks. He takes a piece of chalk and starts writing a sequence of consecutive integers starting with 1 to N (1 < N < 10000) . After that, he counts the number of times each digit (0 to 9) appears in the sequen…

​ Trung is bored with his mathematics homeworks. He takes a piece of chalk and starts writing a sequence of consecutive integers starting with 1 to N (1 < N < 10000). After that, he counts the number of times each digit (0 to 9) appears in the seque…

终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance 44.10% Meidum 475 Heaters  30.20% Easy 474 Ones and Zeroes  34.90% Meidum 473 Matchsticks to Square  31.80% Medium 472 Concatenated Words 29.20% Hard…

Total Accepted: 75767 Total Submissions: 314003 Difficulty: Medium Implement int sqrt(int x). Compute and return the square root of x.  (M) Pow(x, n) class Solution { public: int mySqrt(int x) { ){ return INT_MIN; } ){ ; } ; ; while(low<high){ ; if(m…

otal Accepted: 54356 Total Submissions: 357733 Difficulty: Medium Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 1.除法转换为加法,加法转换为乘法,y个x相加的结果就是x*y. 假设结果为dividend的一半,用此值与divisor相乘,如果相乘结果大于…

请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如:[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down 请下拉滚动条查看最新 Weekly Contest!!! Swift LeetCode 目录 | Catalog 序        号 题名Title 难度     Difficulty  两数之…

突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对应的随笔下面评论区留言,我会及时处理,在此谢过了. 过程或许会很漫长,也很痛苦,慢慢来吧. 编号 题名 过题率 难度 1 Two Sum 0.376 Easy 2 Add Two Numbers 0.285 Medium 3 Longest Substring Without Repeating C…

LeetCode题目解答——Easy部分 Posted on 2014 年 11 月 3 日 by 四火 [Updated on 9/22/2017] 如今回头看来,里面很多做法都不是最佳的,有的从复杂度上根本就不是最优解,有的写的太啰嗦,有的则用了一些过于tricky的方法.我没有为了这个再更新,就让它们去吧. LeetCode最近很火,我以前不太知道有这么一个很方便练习算法的网站,直到大概数周前同事和我说起,正好我老婆要找工作,而根据同事的理论,LeetCode的题目是必须攻破的第一道关卡.…