剑指 Offer 15. 二进制中1的个数693. 交替位二进制数

Posted 2021-06-26 南岸青栀*

剑指 Offer 15. 二进制中1的个数

法1：转换为字符串形式，计数“1”

class Solution:
    def hammingWeight(self, n: int) -> int:
        #法1：转换为字符串形式，计数“1”
        # n = str(bin(n))[2::]
        # print(n)
        # res = 0
        # for s in n:
        #     if s == "1":
        #         res += 1
        # return res 

法2：位运算，依次右移

class Solution:
    def hammingWeight(self, n: int) -> int:

        res = 0
        while n > 0:
            res += (n & 1)
            n = n >> 1
        return res

        #法2：不断循环右移
        res = 0
        while n > 0:
            if n%2 != 0:
                res += 1
            n = n >> 1
        return res

python一行：不解释

class Solution:
    def hammingWeight(self, n: int) -> int:
        return sum(map(int,bin(n)[2::]))

693. 交替位二进制数

法1：位运算

class Solution:
    def hasAlternatingBits(self, n: int) -> bool:
        #右移一位后异或，若为10间隔则全为1
        tmp = n ^ (n >> 1)
        #判断tmp是否全为1
        return tmp&(tmp + 1) == 0

法2：转为二进制字符串判断“11” or “00” 是否存在
存在“11”or“00”则说明不满足

class Solution:
    def hasAlternatingBits(self, n: int) -> bool:
        return not ("11" in str(bin(n))[2::] or "00" in str(bin(n))[2::])