# 693.Binary Number with Alternating Bits

**693.Binary Number with Alternating Bits**

难度:Easy

> 给定一个正整数，检查他是否为交替位二进制数：换句话说，就是他的二进制数相邻的两个位数永不相等。

```
示例 1:

输入: 5
输出: True
解释:
5的二进制数是: 101
示例 2:

输入: 7
输出: False
解释:
7的二进制数是: 111
示例 3:

输入: 11
输出: False
解释:
11的二进制数是: 1011
 示例 4:

输入: 10
输出: True
解释:
10的二进制数是: 1010
```

将该数按位取反，然后移动一位，判断是否相等。

```
class Solution {
public:
    bool hasAlternatingBits(int n) {
        int c= pow(2,int(log2(n))+1)-1;
        return ((n>>1) == (c^n)) || ( (c^n) >>1 == n) ;
    }
};
```


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