lc 338 Counting Bits 338 Counting Bits Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. Example: For num = 5 you should retur…

338.Counting Bits - Medium Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. Example: For num = 5 you should return [0,1,1,2,1…

Counting Bits Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. Example:For num = 5 you should return [0,1,1,2,1,2]. Follow up…

https://leetcode.com/problems/counting-bits/ 给定一个非负数n,输出[0,n]区间内所有数的二进制形式中含1的个数 Example: For num = 5 you should return [0,1,1,2,1,2]. 注意fellow up部分,题目说了你要是一个个无脑去遍历输出是不ok的,直接用某些内置函数也是不行的 解题思路 实在没思路就看看hint部分 找张纸,多写几个数,包括: 1.数(十进制) 2.数(二进制) 3.二进制中1的个数 图…

题目描述: Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. Example:For num = 5 you should return [0,1,1,2,1,2]. Follow up: It is…

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. Example: For num = 5 you should return [0,1,1,2,1,2]. Follow up: It is very…

leetcode是求当前所有数的二进制中1的个数,剑指offer上是求某一个数二进制中1的个数 https://www.cnblogs.com/grandyang/p/5294255.html 第三种方法,利用奇偶性找规律 class Solution { public: vector<int> countBits(int num) { vector<}; ;i <= num;i++){ == ) result.push_back(result[i/]); else result.…

题目: Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. Example: For num = 5 you should return [0,1,1,2,1,2]. Follow up: It is v…

给定一个非负整数 num. 对于范围 0 ≤ i ≤ num 中的每个数字 i ,计算其二进制数中的1的数目并将它们作为数组返回.示例:比如给定 num = 5 ,应该返回 [0,1,1,2,1,2].进阶:    给出时间复杂度为O(n * sizeof(integer)) 的解答非常容易. 但是你可以在线性时间O(n)内用一次遍历做到吗?    要求算法的空间复杂度为O(n).    你能进一步完善解法吗? 在c ++或任何其他语言中不使用任何内置函数(如c++里的 __builtin_po…

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. Example: For num = 5 you should return [0,1,1,2,1,2]. Follow up: It is very…