最近准备刷 leetcode  做到了一个关于位运算的题记下方法 int cunt = 0; while(temp) { temp = temp&(temp - 1);  //把二进制最左边那个1变为零 count++;   //统计1的个数 } 同理把位二进制坐左边那个0变为1 就可以  temp = temp|(temp + 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…

问题描述:给出一个非负整数num,对[0, num]范围内每个数都计算它的二进制表示中1的个数 Example:For num = 5 you should return [0,1,1,2,1,2] 思路:该题属于找规律题,令i从0开始,设f(i)为i对应二进制表示中1的个数,写几对对应值就出来了. 很明显,规律就是[0, 1)部分各个值加1就构成了[1, 2)部分,[0, 2)部分各个值加1就构成了[2, 4)部分,[0, 4)部分各个值加1就构成了[4, 8)部分. 以此类推. 原因也很显然…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 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 a…

1.题目描述 2.问题分析 利用bitset. 3 代码 vector<int> countBits(int num) { vector<int> v; ; i <= num; i++){ bitset<> b(i); v.push_back( b.count() ); } return v; }…

原题 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 ve…

Leetcode之动态规划(DP)专题-338. 比特位计数(Counting Bits) 给定一个非负整数 num.对于 0 ≤ i ≤ num 范围中的每个数字 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 1: 输入: 2 输出: [0,1,1] 示例 2: 输入: 5 输出: [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.但你可以在线性时间O(n)内用一趟扫描做到吗? 要求算法的空间复杂度为O(n). 你能…

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…

原题链接在这里:https://leetcode.com/problems/number-of-1-bits/ 题目: 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 0…

二叉树,结构很简单,只是比单链表复杂了那么一丢丢而已.我们先来看看它们结点上的差异: /* 单链表的结构 */ struct SingleList{ int element; struct SingleList *next; }; /* 二叉树的结构 */ struct BinaryTree{ int element; struct BinaryTree *left; struct BinaryTree *right; }; 根据以上两个结构,我们不难发现,单链表的结点只有一个指向下一结点的指针…