## Thursday, May 10, 2018

### LeetCode 693 - Binary Number with Alternating Bits

https://leetcode.com/problems/binary-number-with-alternating-bits/description/
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.


• Time Complexity: $O(1)$. For arbitrary inputs, we do $O(w)$ work, where $w$ is the number of bits in n. However, $w \leq 32$.
• Space complexity: $O(1)$, or alternatively $O(w)$.
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;
}

public boolean hasAlternatingBits(int n) {
int cur = n % 2;
n /= 2;
while (n > 0) {
if (cur == n % 2) return false;
cur = n % 2;
n /= 2;
}
return true;
}