0693. Binary Number With Alternating Bits

# 693. Binary Number with Alternating Bits#

## 题目 #

Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.

Example 1:

``````Input: 5
Output: True
Explanation:
The binary representation of 5 is: 101
``````

Example 2:

``````Input: 7
Output: False
Explanation:
The binary representation of 7 is: 111.
``````

Example 3:

``````Input: 11
Output: False
Explanation:
The binary representation of 11 is: 1011.
``````

Example 4:

``````Input: 10
Output: True
Explanation:
The binary representation of 10 is: 1010.
``````

## 解题思路 #

• 判断一个数的二进制位相邻两个数是不相等的，即 `0101` 交叉间隔的，如果是，输出 true。这一题有多种做法，最简单的方法就是直接模拟。比较巧妙的方法是通过位运算，合理构造特殊数据进行位运算到达目的。`010101` 构造出 `101010` 两者相互 `&` 位运算以后就为 0，因为都“插空”了。

## 代码 #

``````
package leetcode

// 解法一
func hasAlternatingBits(n int) bool {
/*
n =         1 0 1 0 1 0 1 0
n >> 1      0 1 0 1 0 1 0 1
n ^ n>>1    1 1 1 1 1 1 1 1
n           1 1 1 1 1 1 1 1
n + 1     1 0 0 0 0 0 0 0 0
n & (n+1)   0 0 0 0 0 0 0 0
*/
n = n ^ (n >> 1)
return (n & (n + 1)) == 0
}

// 解法二
func hasAlternatingBits1(n int) bool {
last, current := 0, 0
for n > 0 {
last = n & 1
n = n / 2
current = n & 1
if last == current {
return false
}
}
return true
}

``````

Apr 8, 2023