[Swift]LeetCode869. 從新排序獲得 2 的冪 | Reordered Power of 2

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➤微信公衆號：山青詠芝（shanqingyongzhi）
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Starting with a positive integer N, we reorder the digits in any order (including the original order) such that the leading digit is not zero.

Return true if and only if we can do this in a way such that the resulting number is a power of 2. 

Example 1:

Input: 1
Output: true 

Example 2:

Input: 10
Output: false 

Example 3:

Input: 16
Output: true 

Example 4:

Input: 24
Output: false 

Example 5:

Input: 46
Output: true 

Note:

1. 1 <= N <= 10^9

---

從正整數 N 開始，咱們按任何順序（包括原始順序）將數字從新排序，注意其前導數字不能爲零。

若是咱們能夠經過上述方式獲得 2 的冪，返回 true；不然，返回 false。 

示例 1：

輸入：1
輸出：true

示例 2：

輸入：10
輸出：false

示例 3：

輸入：16
輸出：true

示例 4：

輸入：24
輸出：false

示例 5：

輸入：46
輸出：true 

提示：

1. 1 <= N <= 10^9

---

Runtime: 8 ms

Memory Usage: 18.6 MB

 1 class Solution {
 2     func reorderedPowerOf2(_ N: Int) -> Bool {
 3         var c:Int = counter(N)
 4         for i in 0..<32
 5         {
 6             if counter(1 << i) == c
 7             {
 8                 return true
 9             }
10         }
11         return false
12     }
13 
14     func counter(_ N:Int) ->Int
15     {
16         var N = N
17         var res:Int = 0
18         while(N > 0)
19         {
20             res += Int(pow(10, Double(N % 10)))
21             N /= 10
22         }
23         return res
24     }
25 }

---

12ms

 1 class Solution {
 2     func reorderedPowerOf2(_ N: Int) -> Bool {
 3          let a = count(N);
 4         for i in 0...31 {
 5             if a.elementsEqual(count(1 << i)) {
 6                 return true;
 7             }
 8         }
 9       return false;
10 }
11     
12     func count(_ N: Int) -> [Int] {
13         var n = N
14         var ans =  [Int](repeating: 0, count: 31)
15         while (n > 0) {
16             let indx = n % 10;
17             ans[indx] = ans[indx] + 1;
18             n /= 10;
19         }
20         return ans
21     }
22 }

---

16ms

 1 class Solution {
 2 
 3     let pows2 = [1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912]
 4     var dictArr = [Int: [[Character: Int]]]()
 5 
 6     init() {
 7         for n in pows2 {
 8             let s = String(n)
 9             let val = toStringDict(n)
10             if let arr = dictArr[s.count] {
11                 var varArr = arr
12                 varArr.append(val)
13                 dictArr[s.count] = varArr
14             } else {
15                 dictArr[s.count] = [val]
16             }
17         }
18     }
19 
20     func reorderedPowerOf2(_ N: Int) -> Bool {
21         return dictArr[String(N).count]?.contains(toStringDict(N)) ?? false
22     }
23 
24     func toStringDict(_ n: Int) -> [Character : Int] {
25         let s = String(n)
26         var dict: [Character : Int] = [:]
27         for char in s.characters {
28             if let val = dict[char] {
29                 dict[char] = val + 1
30             } else {
31                 dict[char] = 1
32             }
33         }
34         return dict
35     }
36 }

---

28ms

 1 class Solution {
 2     func reorderedPowerOf2(_ N: Int) -> Bool {
 3     if N == 1 {return true}
 4     var set = [String]()
 5     let stirng = String("\(N)".sorted(by: >))
 6     let max = Int(stirng)!
 7     for i in 1... {
 8         let number = 1 << i
 9         if number > max {break}
10         set.append(String("\(number)".sorted(by: >)))
11     }
12      return set.contains(stirng)
13     }
14 }