[Swift]LeetCode1178. 猜字谜 | Number of Valid Words for Each Puzzle

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With respect to a given puzzle string, a word is valid if both the following conditions are satisfied:

• word contains the first letter of puzzle.
• For each letter in word, that letter is in puzzle.
  For example, if the puzzle is "abcdefg", then valid words are "faced", "cabbage", and "baggage"; while invalid words are "beefed" (doesn't include "a") and "based" (includes "s" which isn't in the puzzle).

Return an array answer, where answer[i] is the number of words in the given word list words that are valid with respect to the puzzle puzzles[i].

 

Example :

Input: 
words = ["aaaa","asas","able","ability","actt","actor","access"], 
puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
Output: [1,1,3,2,4,0]
Explanation:
1 valid word for "aboveyz" : "aaaa" 
1 valid word for "abrodyz" : "aaaa"
3 valid words for "abslute" : "aaaa", "asas", "able"
2 valid words for "absoryz" : "aaaa", "asas"
4 valid words for "actresz" : "aaaa", "asas", "actt", "access"
There're no valid words for "gaswxyz" cause none of the words in the list contains letter 'g'.

 

Constraints:

• 1 <= words.length <= 10^5
• 4 <= words[i].length <= 50
• 1 <= puzzles.length <= 10^4
• puzzles[i].length == 7
• words[i][j], puzzles[i][j] are English lowercase letters.
• Each puzzles[i] doesn't contain repeated characters.

---

 外国友人仿照中国字谜设计了一个英文版猜字谜小游戏，请你来猜猜看吧。

字谜的迷面 puzzle 按字符串形式给出，若是一个单词 word 符合下面两个条件，那么它就能够算做谜底：

• 单词 word 中包含谜面 puzzle 的第一个字母。
• 单词 word 中的每个字母均可以在谜面 puzzle 中找到。
  例如，若是字谜的谜面是 "abcdefg"，那么能够做为谜底的单词有 "faced", "cabbage", 和 "baggage"；而 "beefed"（不含字母 "a"）以及 "based"（其中的 "s" 没有出如今谜面中）。

返回一个答案数组 answer，数组中的每一个元素 answer[i] 是在给出的单词列表 words 中能够做为字谜迷面 puzzles[i] 所对应的谜底的单词数目。

 

示例：

输入：
words = ["aaaa","asas","able","ability","actt","actor","access"], 
puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
输出：[1,1,3,2,4,0]
解释：
1 个单词能够做为 "aboveyz" 的谜底 : "aaaa" 
1 个单词能够做为 "abrodyz" 的谜底 : "aaaa"
3 个单词能够做为 "abslute" 的谜底 : "aaaa", "asas", "able"
2 个单词能够做为 "absoryz" 的谜底 : "aaaa", "asas"
4 个单词能够做为 "actresz" 的谜底 : "aaaa", "asas", "actt", "access"
没有单词能够做为 "gaswxyz" 的谜底，由于列表中的单词都不含字母 'g'。

 

提示：

• 1 <= words.length <= 10^5
• 4 <= words[i].length <= 50
• 1 <= puzzles.length <= 10^4
• puzzles[i].length == 7
• words[i][j], puzzles[i][j] 都是小写英文字母。
• 每一个 puzzles[i] 所包含的字符都不重复。

---

Runtime: 656 ms

Memory Usage: 36.7 MB

 1 class Solution {  2     func findNumOfValidWords(_ words: [String], _ puzzles: [String]) -> [Int] {  3         var result:[Int] = [Int]()  4         var wordSets:[Int:Int] = [Int:Int]()  5         var puzzles = puzzles.map{Array($0).map{$0.ascii}}  6         var words = words.map{Array($0).map{$0.ascii - 97}}  7         // Encode word of 26 chars in 31 bit (signed Int)
 8         for i in 0..<words.count  9  { 10             var bits:Int = 0
11             for j in 0..<words[i].count 12  { 13                 bits |= 1 << words[i][j] 14  } 15             wordSets[bits,default:0] += 1
16  } 17 
18         for i in 0..<puzzles.count 19  { 20             var bits:Int = 0
21             var firstCharBitSet = 1 << (puzzles[i][0] - 97) 22             for j in 0..<puzzles[i].count 23  { 24                 bits |= 1 << (puzzles[i][j] - 97) 25  } 26             var count:Int = 0
27             var b:Int = bits 28             while(b > 0) 29  { 30                 if (b & firstCharBitSet) != 0 && wordSets[b] != nil 31  { 32                     count += wordSets[b]!
33  } 34                 b = --b & bits 35  } 36  result.append(count) 37  } 38         return result 39  } 40 } 41 
42 //Character扩展 
43 extension Character 44 { 45   //Character转ASCII整数值(定义小写为整数值)
46    var ascii: Int { 47        get { 48            return Int(self.unicodeScalars.first?.value ?? 0) 49  } 50  } 51 } 52     
53 /*扩展Int类，实现自增++、自减--运算符*/
54 extension Int{ 55     //--前缀:先自减再执行表达示
56     static prefix func --(num:inout Int) -> Int { 57         //输入输出参数num
58         num -= 1
59         //返回减1后的数值
60         return num 61  } 62 }

---

1020ms

 1 class Solution {  2     func findNumOfValidWords(_ words: [String], _ puzzles: [String]) -> [Int] {  3         let table = Array("abcdefghijklmnopqrstuvwxyz")  4         var dic = [Character: Int](), i = 0
 5         for c in table {  6             dic[c] = i  7             i += 1
 8  }  9         var ans = [Int]() 10         var freq = [Int](repeating: 0, count: 1<<26) 11         for w in words { 12             let cw = Array(w) 13             var mask = 0
14             for c in cw { 15                 mask |= 1<<dic[c]!
16  } 17             freq[mask] += 1
18  } 19         for p in puzzles { 20             let cp = Array(p) 21             var mask = 0
22             for c in cp { 23                 mask |= 1<<dic[c]!
24  } 25 
26             let first = dic[cp.first!]!
27             var sub = mask 28             var total = 0
29             while true { 30                 if sub >> first & 1 != 0 { 31                     total += freq[sub] 32  } 33                 if sub == 0 { 34                     break
35  } 36                 sub = (sub-1)&mask 37  } 38  ans.append(total) 39  } 40         return ans 41  } 42 }

---

1908ms

 1 class Solution {  2     func findNumOfValidWords(_ words: [String], _ puzzles: [String]) -> [Int] {  3         let trie = Trie()  4         var result = [Int]()  5         
 6         for word in words {  7  trie.insert(String(Set(word).sorted()))  8  }  9         
10         for puzzle in puzzles { 11             let firstCharacter = puzzle[String.Index(encodedOffset: 0)] 12             let sortedPuzzle = String(puzzle.sorted()) 13             
14             result.append(trie.search(sortedPuzzle, firstCharacter, false)) 15  } 16         
17         return result 18  } 19 } 20 
21 class TrieNode { 22     var hash = [Character:TrieNode]() 23     var countOfWords = 0
24     
25     func search(_ word: String, _ firstChar: Character, _ firstSeen: Bool) -> Int { 26         var count = 0
27         
28         if firstSeen { 29             count += countOfWords 30  } 31         
32         for i in 0..<word.count { 33             let wordChar = word[String.Index(encodedOffset: i)] 34             
35             guard let childNode = hash[wordChar] else { continue } 36             
37             if wordChar == firstChar { 38                 count += childNode.search(word, firstChar, true) 39             } else { 40                 count += childNode.search(word, firstChar, firstSeen) 41  } 42             
43  } 44         
45         return count 46  } 47 } 48 
49 class Trie { 50     let root = TrieNode() 51     
52     init(_ words: [String] = []) { 53  words.forEach(insert) 54  } 55     
56     func search(_ word: String, _ firstChar: Character, _ firstSeen: Bool) -> Int { 57         return root.search(word, firstChar, firstSeen) 58  } 59     
60  func insert(_ word: String) { 61         var current = root 62         for character in word { 63             if let node = current.hash[character] { 64                 current = node 65             } else { 66                 current.hash[character] = TrieNode() 67                 current = current.hash[character]!
68  } 69  } 70         current.countOfWords += 1
71  } 72 }