[Swift]LeetCode869. 重新排序得到 2 的幂 | Reordered Power of 2_app

概述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 t

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: 1Output: true

Example 2:

input: 10Output: false

Example 3:

input: 16Output: @H_502_37@true

Example 4:

input: 24Output: false

Example 5:

input: 46Output: true

Note:

1 <= N <= 10^9

从正整数 N 开始，我们按任何顺序（包括原始顺序）将数字重新排序，注意其前导数字不能为零。

如果我们可以通过上述方式得到 2 的幂，返回 true；否则，返回 false。

示例 1：

输入：1输出：true

示例 2：

输入：10输出：false

示例 3：

输入：16输出：true

示例 4：

输入：24输出：false

示例 5：

输入：46输出：true

提示：

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 false12     }13 14     func counter(_ N:Int) ->Int15     {16         var N = N17         var res:Int = 018         while(N > 0)19         {20             res += Int(pow(10,Double(N % 10)))21             N /= 1022         }23         return res24     }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 = N14         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 ans21     }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 = arr12                 vararr.append(val)13                 dictArr[s.count] = vararr14             } 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)) ?? false22     }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 + 130             } else {31                 dict[char] = 132             }33         }34         return dict35     }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 }