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[Swift]LeetCode1016. 子串能表示从 1 到 N 数字的二进制串 | Binary String With Substrings Representing 1 To N

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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

给定一个二进制字符串 S(一个仅由若干 '0' 和 '1' 构成的字符串)和一个正整数 N,如果对于从 1 到 N 的每个整数 X,其二进制表示都是 S 的子串,就返回 true,否则返回 false

示例 1:

输入:S = "0110", N = 3
输出:true

示例 2:

输入:S = "0110", N = 4
输出:false

提示:

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

Runtime: 8 ms
Memory Usage: 20.4 MB
 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扩展
21 extension Int
22 {
23     //Int转Character,ASCII值(定义大写为字符值)
24     var ASCII:Character 
25     {
26         get {return Character(UnicodeScalar(self)!)}
27     }
28 }

 

posted @ 2019-03-24 14:19  为敢技术  阅读(446)  评论(0编辑  收藏  举报