693. Binary Number with Alternating Bits Poison 2022-01-19 Bit 123456class Solution { public boolean hasAlternatingBits(int n) { int x = (n ^ (n >>> 1)) + 1; return (x & (x - 1)) == 0; }} 题目的输入数据可以保证不会发生越界,编写代码时注意位运算符号的优先级,如果记不住可以使用括号确保优先运算。 References693. Binary Number with Alternating Bits