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693. 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: n = 5
Output: true
Explanation: The binary representation of 5 is: 101

Example 2:

Input: n = 7
Output: false
Explanation: The binary representation of 7 is: 111.

Example 3:

Input: n = 11
Output: false
Explanation: The binary representation of 11 is: 1011.

 

Constraints:

  • 1 <= n <= 231 - 1

Solutions

Solution 1

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class Solution:
    def hasAlternatingBits(self, n: int) -> bool:
        prev = -1
        while n:
            curr = n & 1
            if prev == curr:
                return False
            prev = curr
            n >>= 1
        return True

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

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class Solution {
public:
    bool hasAlternatingBits(int n) {
        int prev = -1;
        while (n) {
            int curr = n & 1;
            if (prev == curr) return false;
            prev = curr;
            n >>= 1;
        }
        return true;
    }
};

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func hasAlternatingBits(n int) bool {
    prev := -1
    for n != 0 {
        curr := n & 1
        if prev == curr {
            return false
        }
        prev = curr
        n >>= 1
    }
    return true
}

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impl Solution {
    pub fn has_alternating_bits(mut n: i32) -> bool {
        let u = n & 3;
        if u != 1 && u != 2 {
            return false;
        }
        while n != 0 {
            if (n & 3) != u {
                return false;
            }
            n >>= 2;
        }
        true
    }
}

Solution 2

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class Solution:
    def hasAlternatingBits(self, n: int) -> bool:
        n ^= n >> 1
        return (n & (n + 1)) == 0

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class Solution {
    public boolean hasAlternatingBits(int n) {
        n ^= (n >> 1);
        return (n & (n + 1)) == 0;
    }
}

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class Solution {
public:
    bool hasAlternatingBits(int n) {
        n ^= (n >> 1);
        return (n & ((long) n + 1)) == 0;
    }
};

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func hasAlternatingBits(n int) bool {
    n ^= (n >> 1)
    return (n & (n + 1)) == 0
}

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impl Solution {
    pub fn has_alternating_bits(n: i32) -> bool {
        let t = n ^ (n >> 1);
        (t & (t + 1)) == 0
    }
}

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