LWC 53：693. Binary Number with Alternating Bits
传送门：693. Binary Number with Alternating Bits
Problem:
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.
思路：
熟悉JAVA接口的知道，Integer类可以直接把数字转为2进制串。
代码如下：
public boolean hasAlternatingBits(int n) {
String binary = Integer.toBinaryString(n);
char[] cs = binary.toCharArray();
int bit = cs[0] - '0';
for (int i = 1; i < cs.length; ++i) {
if (bit == cs[i] - '0') return false;
bit = cs[i] - '0';
}
return true;
}当然，你也可以自己解析每一位，代码如下：
public boolean hasAlternatingBits(int n) {
int bit = n >> 0 & 1;
n >>= 1;
while (n > 0) {
if (bit == (n & 1)) return false;
bit = n & 1;
n >>= 1;
}
return true;
}或者合并到一块：
public boolean hasAlternatingBits(int n) {
int bit = -1;
while (n > 0) {
if (bit == (n & 1)) return false;
bit = n & 1;
n >>= 1;
}
return true;
}
