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…

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…

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. 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:…

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…

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()…

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. 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…

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…

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…

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…

刷题备忘录,for bug-free leetcode 396. Rotate Function 题意: Given an array of integers A and let n to be its length. Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow: F(k…

刷题备忘录,for bug-free 招行面试题--求无序数组最长连续序列的长度,这里连续指的是值连续--间隔为1,并不是数值的位置连续 问题: 给出一个未排序的整数数组,找出最长的连续元素序列的长度. 如: 给出[100, 4, 200, 1, 3, 2], 最长的连续元素序列是[1, 2, 3, 4].返回它的长度:4. 你的算法必须有O(n)的时间复杂度 . 解法: 初始思路 要找连续的元素,第一反应一般是先把数组排序.但悲剧的是题目中明确要求了O(n)的时间复杂度,要做一次排序,是不能达…

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…

151 Reverse Words in a String class Solution { public: void reverseWords(string &s) { string result; ; i >= ;) { && s[i] == ' ') { i--; } ) { break; } string word; && s[i] != ' ') { word += s[i]; i--; } reverse(word.begin(), word.en…

1. Array 基础 27 Remove Element 26 Remove Duplicates from Sorted Array 80 Remove Duplicates from Sorted Array II 277 Find the Celebrity 189 Rotate Array 41 First Missing Positive 299 Bulls and Cows 134 Gas Station 118 Pascal's Triangle 很少考 119 Pascal's…

LeetCode算法题目解答汇总 本文转自<四火的唠叨> 只要不是特别忙或者特别不方便,最近一直保持着每天做几道算法题的规律,到后来随着难度的增加,每天做的题目越来越少.我的初衷就是练习,因为一方面我本身算法基础并不好,再一方面是因为工作以后传统意义上所谓算法的东西接触还是太少.为了题目查找方便起见,我把之前几篇陆陆续续贴出来的我对LeetCode上面算法题的解答汇总在下面,CTRL+F就可以比较方便地找到.由于LeetCode上的题在不断更新,因此我也会不定期地更新.下面表格里面的Accep…

先过一下Hard模式的题目吧.   # Title Editorial Acceptance Difficulty Frequency   . 65 Valid Number     12.6% Hard    . 126 Word Ladder II     13.6% Hard    . 149 Max Points on a Line     15.6% Hard    . 146 LRU Cache     16.0% Hard    . 68 Text Justification  …

接上一篇:http://www.cnblogs.com/charlesblc/p/6283064.html 继续过Hard模式的题目吧.   # Title Editorial Acceptance Difficulty Frequency   . 65 Valid Number     12.6% Hard    . 126 Word Ladder II     13.6% Hard    . 149 Max Points on a Line     15.6% Hard    . 146 L…

http://www.cnblogs.com/charlesblc/p/6372971.html 继续过Hard模式的题目吧.   # Title Editorial Acceptance Difficulty Frequency   . 65 Valid Number     12.6% Hard    . 126 Word Ladder II     13.6% Hard    . 149 Max Points on a Line     15.6% Hard    . 146 LRU Ca…

接上一篇:http://www.cnblogs.com/charlesblc/p/6364102.html 继续过Hard模式的题目吧.   # Title Editorial Acceptance Difficulty Frequency   . 65 Valid Number     12.6% Hard    . 126 Word Ladder II     13.6% Hard    . 149 Max Points on a Line     15.6% Hard    . 146 L…

http://www.cnblogs.com/charlesblc/p/6384132.html 继续过Hard模式的题目吧.   # Title Editorial Acceptance Difficulty Frequency   . 65 Valid Number     12.6% Hard    . 126 Word Ladder II     13.6% Hard    . 149 Max Points on a Line     15.6% Hard    . 146 LRU Ca…

http://www.cnblogs.com/charlesblc/p/6384132.html 继续过Hard模式的题目吧.   # Title Editorial Acceptance Difficulty Frequency   . 65 Valid Number     12.6% Hard    . 126 Word Ladder II     13.6% Hard    . 149 Max Points on a Line     15.6% Hard    . 146 LRU Ca…

  # Title Editorial Acceptance Difficulty Frequency   . 65 Valid Number     12.6% Hard    . 126 Word Ladder II     13.6% Hard    . 149 Max Points on a Line     15.6% Hard    . 146 LRU Cache     16.0% Hard    . 68 Text Justification     18.1% Hard    …