LeetCode 693. Binary Number with Alternating Bits

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: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.

代码实现如下：

思路1：

用 t 表示该数字的最后一位，比如10的二进制为1010，则 t = 0;

n /= 2; 如果其二进制最后一位 t 总是等于 n / 2 后剩下的最后一位，说明其二进制相邻位置有相同值，即不交互，返回false;

否则返回true；

class Solution {

    public boolean hasAlternatingBits(int n) {

        int t = n % 2;

        n /= 2;

        while(n > 0){

            if(t == n % 2) return false;

            t = n % 2;

            n /= 2;

        }

        return true;

    }

}

思路2：

调用Integer.toBinaryString()方法将十进制数转换位二进制字符串，依次比较当前位是否与下一位相等，相等返回false,否则返回true

class Solution {

    public boolean hasAlternatingBits(int n) {

        String bits = Integer.toBinaryString(n);

        for(int i = 0; i < bits.length() - 1; i++){

            if(bits.charAt(i) == bits.charAt(i + 1)) return false;

        }

        return true;

    }

}