RUST GYM Leetcode 693 Binary Number with Alternating Bits Rust Solution

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: 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.

Example 4:

Input: n = 10
Output: true
Explanation: The binary representation of 10 is: 1010.

Example 5:

Input: n = 3
Output: false

Constraints:

1 <= n <= 2³¹ - 1

Rust Solution

struct Solution;

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

#[test]
fn test() {
    assert_eq!(Solution::has_alternating_bits(5), true);
    assert_eq!(Solution::has_alternating_bits(7), false);
    assert_eq!(Solution::has_alternating_bits(11), false);
    assert_eq!(Solution::has_alternating_bits(1), true);
}

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