[Swift]LeetCode693. 交替位二进制数 | Binary Number with Alternating Bits

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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.

---

给定一个正整数，检查他是否为交替位二进制数：换句话说，就是他的二进制数相邻的两个位数永不相等。

示例 1:

输入: 5
输出: True
解释:
5的二进制数是: 101

示例 2:

输入: 7
输出: False
解释:
7的二进制数是: 111

示例 3:

输入: 11
输出: False
解释:
11的二进制数是: 1011

 示例 4:

输入: 10
输出: True
解释:
10的二进制数是: 1010

---

Runtime: 4 ms

Memory Usage: 18.4 MB

class Solution {
    func hasAlternatingBits(_ n: Int) -> Bool {
        return ((n + (n >> 1) + 1) & (n + (n >> 1))) == 0
    }
}

---

4ms

class Solution {
    func hasAlternatingBits(_ n: Int) -> Bool {
        
        var current = n % 2
        var n = n / 2
        while n > 0 {
            if current == n % 2 {
                return false
            }
            current = n % 2
            n = n / 2
        }
        return true
    }
}

---

12ms

class Solution {
    func hasAlternatingBits(_ n: Int) -> Bool {
        let binaryString = String(n, radix: 2)
        let count = binaryString.count
        var i = 0
        var flag = -1
        while i < binaryString.count {
            if flag != -1 {
                guard flag != (n>>i) & 1 else {
                    return false
                }
            }
            flag = (n>>i) & 1
            i += 1
        }
        return true        
    }
}

---

12ms

class Solution {
    func hasAlternatingBits(_ n: Int) -> Bool {
        let str = String(n,radix:2)
        let characters = Array(str)
        var result = true
        for i in 0..<characters.count {
            if i+1 < characters.count {
                if characters[i] == characters[i+1] {
                    result = false
                }
            }
        }
        return result
    }
}

---

24ms

class Solution {
    func hasAlternatingBits(_ n: Int) -> Bool {
        var n = n
        var bit0 = n & 1
        n >>= 1
        while n > 0 {
            if n & 1 != bit0 {
                bit0 = n & 1
                n >>= 1
                continue
            } else {
                return false
            }
        }
        return true
    }
}

---

28ms

class Solution {
    func hasAlternatingBits(_ n: Int) -> Bool {
       let str = toBinary(n)
        for i in 0 ..< str.count - 1 {
            let index1 = str.index(str.startIndex, offsetBy: i)
            let index2 = str.index(str.startIndex, offsetBy: i + 1)
            let c = str[index1 ..< index2]
            let index3 = str.index(str.startIndex, offsetBy: i + 2)
            let c1 = str[index2 ..< index3]
            if String(c) == String(c1) {
                return false
            }
        }
        return true
    }
    
    func toBinary(_ n: Int) -> String {
        var m = n
        var str = ""
        while m > 0 {
            let temp = m % 2
            str = "\(temp)" + str
            m /= 2
        }
        return str
    }
}