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一: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 00000000000000000000000000001011, so t…
Write a function that takes an unsigned integer and return the number of '1' bits it has (also known as the Hamming weight). Example 1: Input: 00000000000000000000000000001011 Output: 3 Explanation: The input binary string 000000000000000000000000000…
编写一个函数,输入是一个无符号整数,返回其二进制表达式中数字位数为 ‘1’ 的个数(也被称为汉明重量) 示例 1: 输入:00000000000000000000000000001011 输出:3 解释:输入的二进制串 00000000000000000000000000001011 中,共有三位为 '1'. 示例 2: 输入:00000000000000000000000010000000 输出:1 解释:输入的二进制串 00000000000000000000000010000000 中,共…
1.题目 2.思路 方法一:常规方法. 方法二:给面试官惊喜的解法. 3.java代码 方法一代码: public class Solution { // you need to treat n as an unsigned value public int hammingWeight(int n) { int count=0; int flag=1; while(flag!=0){ if((n&flag)!=0) count++; flag=flag<<1; } return cou…
public class Solution { public int HammingWeight(uint n) { var list = new List<uint>(); do { ; list.Add(x); n = n / ; } ); ; ; i < list.Count; i++) { ) { count++; } } return count; } } https://leetcode.com/problems/number-of-1-bits/#/description…
题意:一个int类型正整数,求它的二进制形式有多少个1 思路:除2递归,可以解出,看了discuss里面有个解更牛,一行结束战斗,是用n&(n-1)再递归,其实并不是很懂怎么想出来这么做的,可能是自己对二进制的处理根本不怎么了解吧,但是这样做结果是对的 代码: int hammingWeight(uint32_t n) { ) ; == ) + hammingWeight(n/); else ); } 大神代码是这样的: int hammingWeight(uint32_t n) { ? : +…
位运算相关 三道题 231. Power of Two Given an integer, write a function to determine if it is a power of two. (Easy) 分析: 数字相关题有的可以考虑用位运算,例如&可以作为筛选器. 比如n & (n - 1) 可以将n的最低位1删除,所以判断n是否为2的幂,即判断(n & (n - 1) == 0) 代码: class Solution { public: bool isPowerOf…
按位操作符只能用于整数基本数据类型中的单个bit中,操作符对应表格: Operator Description & 按位与(12345&1=1,可用于判断整数的奇偶性) | 按位或 ^ 异或(同假异真) ~ 非(一元操作符) &=,|=,^= 合并运算和赋值 <<N 左移N位,低位补0 >>N 右移N位,(正数:高位补0,负数高位补1) >>>N 无符号右移(无论正负数,高位皆补0) <<=,>>=,>>…