题解 | #二进制中1的个数#
二进制中1的个数
https://www.nowcoder.com/practice/8ee967e43c2c4ec193b040ea7fbb10b8?tpId=13&tqId=23273&ru=/exam/oj/ta&qru=/ta/coding-interviews/question-ranking&sourceUrl=%2Fexam%2Foj%2Fta%3Fpage%3D1%26tpId%3D13%26type%3D13
# # 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 # # # @param n int整型 # @return int整型 # class Solution: def NumberOf1(self , n: int) -> int: # write code here x=1 res=0 for i in range(32): if (n & (1<<i)) !=0: res+=1 return res