Java

  • /**
    
     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.
    
     */
    public class Binary_Number_with_Alternating_Bits {
    
        public static void main (String[] args) {
            Binary_Number_with_Alternating_Bits out = new Binary_Number_with_Alternating_Bits();
            Solution s = out.new Solution();
    
            System.out.println(s.hasAlternatingBits(3)); // 11
            System.out.println(s.hasAlternatingBits(4)); // 100
            System.out.println(s.hasAlternatingBits(5)); // 101
        }
    
        class Solution {
            public boolean hasAlternatingBits(int n) {
    
                if (n <= 0) {
                    return false;
                }
    
                int prev = n & 1; // true => 1, false => 0
                n >>= 1;
    
                while (n > 0) {
    
                    if (prev == (n & 1)) {
                        return false;
                    }
    
                    prev = n & 1;
                    n >>= 1;
                }
    
                return true;
            }
        }
    }
    
  • // OJ: https://leetcode.com/problems/binary-number-with-alternating-bits/
    // Time: O(1)
    // Space: O(1)
    class Solution {
    public:
        bool hasAlternatingBits(int n) {
            int prev = -1;
            while (n) {
                int d = n % 2;
                n /= 2;
                if (prev == d) return false;
                prev = d;
            }
            return true;
        }
    };
    
  • In [1]: bin(5)
    Out[1]: '0b101'
    

Java

  • class Solution {
        public boolean hasAlternatingBits(int n) {
            if (n <= 2)
                return true;
            else if (n <= 4)
                return false;
            int prevBit = n & 1;
            n >>= 1;
            while (n > 0) {
                int curBit = n & 1;
                if (prevBit == curBit)
                    return false;
                prevBit = curBit;
                n >>= 1;
            }
            return true;
        }
    }
    
  • // OJ: https://leetcode.com/problems/binary-number-with-alternating-bits/
    // Time: O(1)
    // Space: O(1)
    class Solution {
    public:
        bool hasAlternatingBits(int n) {
            int prev = -1;
            while (n) {
                int d = n % 2;
                n /= 2;
                if (prev == d) return false;
                prev = d;
            }
            return true;
        }
    };
    
  • In [1]: bin(5)
    Out[1]: '0b101'
    

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