原题链接：leetcode 693. Binary Number with Alternating Bits
Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.
Example 1:
Input: n = 5 Output: true Explanation: The binary representation of 5 is: 101
Example 2:
Input: n = 7 Output: false Explanation: The binary representation of 7 is: 111.
Example 3:
Input: n = 11 Output: false Explanation: The binary representation of 11 is: 1011.
Constraints:
1 <= n <= 231 - 1
做法一：模拟 思路：
对于n，每次让n对2取余求出当前位，和上一位进行对比，如果相同false，循环结束返回true
C++代码：class Solution {
public:
bool hasAlternatingBits(int n) {
// pre 上一位
int pre = 2;
while(n){
// cnt 当前位
int cnt = n % 2;
if(cnt == pre) return false;
else{
pre = cnt;
n /= 2;
}
}
return true;
}
};
python3代码：
class Solution:
def hasAlternatingBits(self, n: int) -> bool:
pre = 2
while n:
cnt = n % 2
if cnt == pre: return False
else:
pre = cnt
n //= 2
return True
复杂度分析：
时间复杂度：O(logn)，n的二进制表示最多logn位，因此循环最多执行logn次空间复杂度：O(1)，用了一个临时变量pre
