LWC 53：693. Binary Number with Alternating Bits

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;    }