Python—位运算的运用 判断奇偶 x & 1 取半 x >> 1 翻倍 x << 1 异或 ^ # 相同为0,不同为1 x ^ 0 = x x ^ x = 0 2的幂次方 x & (x - 1) = 0 x二进制中1的个数(Brian Kernighan 算法) count = 0 while x != 0: x = x & (x - 1) count += 1 二分查找中mid获取 mid = low + ((high - low) >> 1)