[LeetCode] 693. Binary Number with Alternating Bits

轉自LeetCode

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.

<Solution>

想法如下

• 先找出 LSB 是 1 還是 0，然後一個一個 bit 檢查是否是預期的答案。例如 LSB 是 0 的話，下一個 bit 就應該是 1，再下一個 bit 是 0

code 如下

Java

	class Solution {
	public boolean hasAlternatingBits(int n) {
	int status = n & 1;
	while(n != 0) {
	if((n & 1) != status) {
	return false;
	}
	n >>= 1;
	status = 1 - status;
	}
	return true;
	}
	}

view raw hasAlternatingBits.java hosted with ❤ by GitHub

Java 還可以利用一些內建函式來解

code 如下

	class Solution {
	public boolean hasAlternatingBits(int n) {
	String s = Integer.toBinaryString(n);
	final int len = s.length();
	for(int i = 1; i < s.length(); i++) {
	if(s.charAt(i) == s.charAt(i-1)) {
	return false;
	}
	}
	return true;
	}
	}

view raw hasAlternatingBits_2.java hosted with ❤ by GitHub