[Swift]LeetCode693.交替位二进制数|BinaryNumberwithAlternatingBits

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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

1 class Solution {
2     func hasAlternatingBits(_ n: Int) -> Bool {
3         return ((n + (n >> 1) + 1) & (n + (n >> 1))) == 0
4     }
5 }

4ms

 1 class Solution {
 2     func hasAlternatingBits(_ n: Int) -> Bool {
 3         
 4         var current = n % 2
 5         var n = n / 2
 6         while n > 0 {
 7             if current == n % 2 {
 8                 return false
 9             }
10             current = n % 2
11             n = n / 2
12         }
13         return true
14     }
15 }

12ms

 1 class Solution {
 2     func hasAlternatingBits(_ n: Int) -> Bool {
 3         let binaryString = String(n, radix: 2)
 4         let count = binaryString.count
 5         var i = 0
 6         var flag = -1
 7         while i < binaryString.count {
 8             if flag != -1 {
 9                 guard flag != (n>>i) & 1 else {
10                     return false
11                 }
12             }
13             flag = (n>>i) & 1
14             i += 1
15         }
16         return true        
17     }
18 }

12ms

 1 class Solution {
 2     func hasAlternatingBits(_ n: Int) -> Bool {
 3         let str = String(n,radix:2)
 4         let characters = Array(str)
 5         var result = true
 6         for i in 0..<characters.count {
 7             if i+1 < characters.count {
 8                 if characters[i] == characters[i+1] {
 9                     result = false
10                 }
11             }
12         }
13         return result
14     }
15 }

24ms

 1 class Solution {
 2     func hasAlternatingBits(_ n: Int) -> Bool {
 3         var n = n
 4         var bit0 = n & 1
 5         n >>= 1
 6         while n > 0 {
 7             if n & 1 != bit0 {
 8                 bit0 = n & 1
 9                 n >>= 1
10                 continue
11             } else {
12                 return false
13             }
14         }
15         return true
16     }
17 }

28ms

 1 class Solution {
 2     func hasAlternatingBits(_ n: Int) -> Bool {
 3        let str = toBinary(n)
 4         for i in 0 ..< str.count - 1 {
 5             let index1 = str.index(str.startIndex, offsetBy: i)
 6             let index2 = str.index(str.startIndex, offsetBy: i + 1)
 7             let c = str[index1 ..< index2]
 8             let index3 = str.index(str.startIndex, offsetBy: i + 2)
 9             let c1 = str[index2 ..< index3]
10             if String(c) == String(c1) {
11                 return false
12             }
13         }
14         return true
15     }
16     
17     func toBinary(_ n: Int) -> String {
18         var m = n
19         var str = ""
20         while m > 0 {
21             let temp = m % 2
22             str = "\(temp)" + str
23             m /= 2
24         }
25         return str
26     }
27 }