Given a binary string S (a string consisting only of ‘0‘ and ‘1‘s) and a positive integer N, return true if and only if for every integer X from 1 to N, the binary representation of X is a substring of S.

Example 1:

Input: S = "0110", N = 3
Output: true

Example 2:

Input: S = "0110", N = 4
Output: false
 

Note:

1. 1 <= S.length <= 1000
2. 1 <= N <= 10^9

給定一個二進制字符串 S（一個僅由若幹 ‘0‘ 和 ‘1‘ 構成的字符串）和一個正整數 N，如果對於從 1 到 N 的每個整數 X，其二進制表示都是 S 的子串，就返回 true，否則返回 false。

示例 1：

輸入：S = "0110", N = 3
輸出：true

示例 2：

輸入：S = "0110", N = 4
輸出：false

提示：

1. 1 <= S.length <= 1000
2. 1 <= N <= 10^9

 1
 
 class Solution {
 2     func queryString(_ S: String, _ N: Int) -> Bool {
 3         if N > 2400 {return false}
 4         for i in 1...N
 5         {
 6             var str:String = String()
 7             var x:Int = i
 8             while (x != 0)
 9             {
10                 str.append((x % 2
 
 + 48).ASCII)
11                 x /= 2
12             }
13             str = String(str.reversed())
14             if !S.contains(str) {return false}
15         }
16         return true        
17     }
18 }
19 
20 //Int擴展
21 extension Int
22 {
23     //Int轉Character,ASCII值(定義大寫為字符值)
24     var ASCII:Character 
25     {
26         get {return Character(UnicodeScalar(self)!)}
27     }
28 }

[Swift Weekly Contest 129]LeetCode1023. 子串能表示從 1 到 N 數字的二進制串 | Binary String With Substrings Representing 1 To N