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.…

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. 大…

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…

18.4 Write a method to count the number of 2s between 0 and n. 这道题给了我们一个整数n,让我们求[0,n]区间内所有2出现的个数,比如如果n=20,那么满足题意的是2, 12, 20,那么返回3即可.LeetCode上有一道很类似的题Factorial Trailing Zeroes,但是那道题求5的个数还包括了因子中的5,比如10里面也有5,这是两题的不同之处.那么首先这题可以用brute force来解,我们对区间内的每一个数字…

Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight). For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should retu…

Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Count the number of distinct island…

题意: 提供一个无符号32位整型uint32_t变量n,返回其二进制形式的1的个数. 思路: 考察二进制的特性,设有k个1,则复杂度为O(k).考虑将当前的数n和n-1做按位与,就会将n的最后一个1去掉,重复这样的操作就可以统计出1的个数了.(2015年春季 小米实习生的笔试题之一) class Solution { public: int hammingWeight(uint32_t n) { ; while(n) { n&=n-; cnt++; } return cnt; } }; AC代码…

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…

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…