zoukankan      html  css  js  c++  java
  • [Swift]LeetCode269. 外星人词典 $ Alien Dictionary

    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
    ➤微信公众号:山青咏芝(shanqingyongzhi)
    ➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
    ➤GitHub地址:https://github.com/strengthen/LeetCode
    ➤原文地址: https://www.cnblogs.com/strengthen/p/10640352.html 
    ➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
    ➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

    There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

    Example 1:

    Input:
    [
      "wrt",
      "wrf",
      "er",
      "ett",
      "rftt"
    ]
    
    Output: "wertf"
    

    Example 2:

    Input:
    [
      "z",
      "x"
    ]
    
    Output: "zx"
    

    Example 3:

    Input:
    [
      "z",
      "x",
      "z"
    ] 
    
    Output: "" 
    
    Explanation: The order is invalid, so return "".
    

    Note:

    1. You may assume all letters are in lowercase.
    2. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary.
    3. If the order is invalid, return an empty string.
    4. There may be multiple valid order of letters, return any one of them is fine.

    有一种新的外来语言使用拉丁字母。 但是,您不知道字母之间的顺序。 您会从字典中收到一个非空单词列表,其中的单词按照新语言的规则按字典顺序排序。 导出这种语言的字母顺序。

    例1:

    Input:
    [
      "wrt",
      "wrf",
      "er",
      "ett",
      "rftt"
    ]
    
    Output: "wertf"

    例2:

    Input:
    [
      "z",
      "x"
    ]
    
    Output: "zx"

    例3:

    Input:
    [
      "z",
      "x",
      "z"
    ] 
    
    Output: "" 

    说明:订单无效,因此请返回“”。
    注意:

    1、您可以假设所有字母都是小写的。
    2、您可以假设如果a是b的前缀,则a必须出现在给定字典中的b之前。
    3、如果订单无效,则返回空字符串。
    4、可能有多个有效的字母顺序,返回其中任何一个都没问题。


    Solution

     1 class Solution {
     2     func alienOrder(_ words: [String]) -> String {
     3         var st:Set<[Character]> = Set<[Character]>()
     4         var ch:Set<Character> = Set<Character>()
     5         var ins:[Int] = [Int](repeating:0,count:256)
     6         var q:[Character] = [Character]()
     7         var res:String = String()
     8         for a in words
     9         {
    10             for c in a
    11             {
    12                 ch.insert(c)
    13             }
    14         }
    15         for i in 0..<words.count - 1
    16         {
    17             let mn:Int = min(words[i].count, words[i + 1].count)
    18             var j:Int = 0
    19             while(j < min(words[i].count, words[i + 1].count))
    20             {
    21                 if words[i][j] != words[i + 1][j]
    22                 {
    23                     st.insert([words[i][j], words[i + 1][j]])
    24                     break
    25                 }
    26                 j += 1
    27             }
    28             if j == mn && words[i].count > words[i + 1].count
    29             {
    30                 return String()
    31             }
    32         }
    33         for a in st
    34         {
    35             ins[a[1].ascii] += 1
    36         }
    37         for a in ch
    38         {
    39             if ins[a.ascii] == 0
    40             {
    41                 q.append(a)
    42                 res.append(a)
    43             }
    44         }
    45         while(!q.isEmpty)
    46         {
    47             let c:Character = q.removeFirst()
    48             for a in st
    49             {
    50                 if a[0] == c
    51                 {
    52                     ins[a[1].ascii] -= 1
    53                     if ins[a[1].ascii] == 0
    54                     {
    55                         q.append(a[1])
    56                         res.append(a[1])
    57                     }
    58                 }
    59             }
    60         }
    61         return res.count == ch.count ? res : String()
    62     }
    63 }
    64 
    65 //String扩展
    66 extension String {
    67     //subscript函数可以检索数组中的值
    68     //直接按照索引方式截取指定索引的字符
    69     subscript (_ i: Int) -> Character {
    70         //读取字符
    71         get {return self[index(startIndex, offsetBy: i)]}
    72     }
    73 }
    74 
    75 //Character扩展
    76 extension Character
    77 {
    78     //Character转ASCII整数值(定义小写为整数值)
    79     var ascii: Int {
    80         get {
    81             return Int(self.unicodeScalars.first?.value ?? 0)
    82         }
    83     }
    84 }

    点击:Playground测试

    1 var sol = Solution()
    2 print(sol.alienOrder(["wrt", "wrf", "er", "ett","rftt"]))
    3 //Print wertf
    4 print(sol.alienOrder(["z","x"]))
    5 //Print zx
    6 print(sol.alienOrder(["z","x","z"]))
    7 //Print ""
  • 相关阅读:
    百度云推送
    web请求报出 “超过了最大请求长度” 【注意:重启IIS】
    页面多个Jquery版本共存的冲突问题,解决方法!
    Web Api 中使用 PCM TO WAV 的语音操作
    Web Api 如何做上传文件的单元测试
    那些年收集的前端学习资源
    原创: 做一款属于自己风格的音乐播放器 (HTML5的Audio新特性)
    Web Api 接口文档制作
    如何在Asp.Net WebApi接口中,验证请求参数中是否携带token标识!
    JavaScript 面试题,给大家补补基础,加加油,埋埋坑!
  • 原文地址:https://www.cnblogs.com/strengthen/p/10640352.html
Copyright © 2011-2022 走看看