为有牺牲多壮志，敢教日月换新天。

[Swift]LeetCode916.单词子集 | Word Subsets

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/9857027.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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

We are given two arrays A and B of words.  Each word is a string of lowercase letters.

Now, say that word b is a subset of word a if every letter in b occurs in a, including multiplicity.  For example, "wrr"is a subset of "warrior", but is not a subset of "world".

Now say a word a from A is universal if for every b in B, b is a subset of a. 

Return a list of all universal words in A.  You can return the words in any order.

Example 1:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
Output: ["facebook","google","leetcode"]

Example 2:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
Output: ["apple","google","leetcode"]

Example 3:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
Output: ["facebook","google"]

Example 4:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
Output: ["google","leetcode"]

Example 5:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
Output: ["facebook","leetcode"]

Note:

  1. 1 <= A.length, B.length <= 10000
  2. 1 <= A[i].length, B[i].length <= 10
  3. A[i] and B[i] consist only of lowercase letters.
  4. All words in A[i] are unique: there isn't i != j with A[i] == A[j].

我们给出了两个数组A和B单词。每个单词都是一串小写字母。

现在，假设每个字母都出现在，包括多重性，那么单词b就是单词的一个子集。例如，"wrr"是"warrior"的子集，但不是"world"的子集。

现在说一个字a从A是普遍的，若对所有b的B，b 是的一个子集a。 

返回所有通用单词的列表A。您可以按任何顺序返回单词。

例1：

输入： A = [“amazon”，“apple”，“facebook”，“google”，“leetcode”]，B = [“e”，“o”] 
输出：[“facebook”，“google”，“leetcode “]

例2：

输入： A = [“amazon”，“apple”，“facebook”，“google”，“leetcode”]，B = [“l”，“e”] 
输出：[“apple”，“google”，“leetcode “]

例3：

输入： A = [“amazon”，“apple”，“facebook”，“google”，“leetcode”]，B = [“e”，“oo”] 
输出：[“facebook”，“google”]

例4：

输入： A = [“amazon”，“apple”，“facebook”，“google”，“leetcode”]，B = [“lo”，“eo”] 
输出：[“google”，“leetcode”]

例5：

输入： A = [“amazon”，“apple”，“facebook”，“google”，“leetcode”]，B = [“ec”，“oc”，“ceo”] 
输出：[“facebook”，“leetcode “]

注意：

  1. 1 <= A.length, B.length <= 10000
  2. 1 <= A[i].length, B[i].length <= 10
  3. A[i]并且B[i]仅由小写字母组成。
  4. 在所有的话A[i]都是独一无二的：没有i != j用A[i] == A[j]。
1536ms
 1 class Solution {
 2     func wordSubsets(_ A: [String], _ B: [String]) -> [String] {
 3         var patterns = Dictionary(B[0].map { ($0, 1) }, uniquingKeysWith: +)
 4         for b in B {
 5             updatePatterns(&patterns, b)
 6         }
 7         var results = [String]()
 8         outer: for a in A {
 9             let dict = Dictionary(a.map { ($0, 1) }, uniquingKeysWith: +)
10             for key in patterns.keys {
11                 guard let value = dict[key] else {
12                     continue outer
13                 }
14                 if value < patterns[key]! {
15                     continue outer
16                 }
17             }
18             results.append(a)
19         }
20         return results
21     }
22     
23     fileprivate func updatePatterns(_ patterns: inout [Character : Int], _ library: String) {
24         var chars = Array(library)
25         let dict = Dictionary(chars.map { ($0, 1) }, uniquingKeysWith: +) 
26         for key in dict.keys {
27             if let pattern = patterns[key] {
28                 patterns[key] = max(pattern, dict[key]!)
29             } else {
30                 patterns[key] = dict[key]
31             }
32         }
33     }
34 }

1864ms

 1 class Solution {
 2     func wordSubsets(_ A: [String], _ B: [String]) -> [String] {
 3         var bmax:[Int] = [Int](repeating: 0,count: 26)
 4         for b in B
 5         {
 6             var bCount:[Int] = count(b)
 7             for i in 0..<26
 8             {
 9                 bmax[i] = max(bmax[i], bCount[i])
10             }
11         }
12         var ans:[String] = [String]()
13         //类似goto语句功能
14         search: 
15         for a in A
16         {
17             var aCount:[Int] = count(a)
18             for i in 0..<26
19             {
20                 if aCount[i] < bmax[i]
21                 {
22                      
23                     continue search
24                    
25                 }  
26             }
27              ans.append(a)
28         }
29         return ans
30     }
31     
32     func count(_ str:String) -> [Int]
33     {
34         var ans:[Int] = [Int](repeating: 0,count: 26)
35         for char in str.characters
36         {
37             //A:65 a:97
38             ans[char.toInt() - 97] += 1 
39         }
40         return ans
41     }
42 }
43 
44 //Character扩展方法  
45 extension Character  
46 {  
47     func toInt() -> Int  
48     {  
49         var num:Int = Int()
50         for scalar in String(self).unicodeScalars  
51         {  
52             num = Int(scalar.value)  
53         }  
54         return num  
55     }  
56 }

2056ms

 1 class Solution {
 2     func wordSubsets(_ A: [String], _ B: [String]) -> [String] {
 3         var hms: [Character:Int] = [:]
 4         for b in B {
 5             var locHm: [Character:Int] = [:]
 6             for symbol in b {
 7                 let temp = (locHm[symbol] ?? 0) + 1
 8                 locHm[symbol] = temp
 9                 hms[symbol] = max((hms[symbol] ?? 0), temp)
10             }
11         }
12         
13         var res: [Int] = []
14         var aHms: [[Character:Int]] = Array(repeating: [:], count: A.count)
15         for (index, a) in A.enumerated() {
16             for symbol in a {
17                 aHms[index][symbol] = (aHms[index][symbol] ?? 0) + 1
18              }
19         }
20         
21          for (index, hm) in aHms.enumerated() {
22              var ok = true
23             for k in hms.keys {
24                 if (hm[k] ?? 0) < hms[k]! {
25                     ok = false
26                     break
27                 }
28             }
29              if ok {
30                  res.append(index)
31              }
32          }
33         
34         return res.map({A[$0]})
35     }
36 }

2208ms

 1 class Solution {
 2     func wordSubsets(_ A: [String], _ B: [String]) -> [String] {
 3         let bDicts:[[Character:Int]] = B.map { wordDict($0) }
 4         var bWord: [Character:Int] = [:]
 5         for bDict in bDicts {
 6             for (ch, count) in bDict {
 7                 bWord[ch] = max((bWord[ch] ?? 0), count)
 8             }
 9         }
10         
11         var result: [String] = []
12         for aWord in A {
13             let aDict = wordDict(aWord)
14             if contains(aDict, bWord) {
15                 result.append(aWord)
16             }
17         }
18         return result
19     }
20     
21     func wordDict(_ str: String) -> [Character:Int] {
22         var result: [Character:Int] = [:]
23         for ch in str {
24             result[ch] = (result[ch] ?? 0) + 1
25         }
26         return result
27     }
28     
29     func contains(_ aDict:[Character:Int], _ bDict:[Character: Int]) -> Bool {
30         for (key, count) in bDict {
31             if aDict[key] == nil || aDict[key]! < count {
32                 return false
33             }
34         }
35         return true
36     }
37 }

2240ms

 1 class Solution {
 2     func wordSubsets(_ A: [String], _ B: [String]) -> [String] {
 3         let universe: [Character: Int] = B.reduce(into: [:]) { res, next in
 4             let c = countCharacters(in: next)
 5             for kv in c {
 6                 res[kv.key] = max(kv.value, res[kv.key] ?? 0)
 7             }
 8         }
 9         return A.filter { self.isWord($0, universalIn: universe) }
10     }
11     
12     func isWord(_ word: String, universalIn universe: [Character: Int]) -> Bool {
13         var wordCount = countCharacters(in: word)
14         for kv in universe {
15             guard let count = wordCount[kv.key], count >= kv.value else {
16                 return false
17             }
18         }
19         return true
20     }
21     
22     func countCharacters(in word: String) -> [Character: Int] {
23         return word.reduce(into: [:]) { (result, next) in
24             let count = result[next] ?? 0
25             result[next] = count + 1
26         }
27     }
28 }

2484ms

 1 class Solution {
 2     func wordSubsets(_ A: [String], _ B: [String]) -> [String] {
 3         let universe: [Character: Int] = B.reduce(into: [:]) { res, next in
 4             let c = countCharacters(in: next)
 5             for kv in c {
 6                 res[kv.key] = max(kv.value, res[kv.key] ?? 0)
 7             }
 8         }
 9         
10         return A.filter { (word) -> Bool in
11             return self.isWord(word, universalIn: universe)
12         }
13     }
14     
15     func isWord(_ word: String, universalIn universe: [Character: Int]) -> Bool {
16         var wordCount = countCharacters(in: word)
17         for kv in universe {
18             guard let count = wordCount[kv.key], count >= kv.value else {
19                 return false
20             }
21         }
22         return true
23     }
24     
25     func countCharacters(in word: String) -> [Character: Int] {
26         return word.reduce(into: [:]) { (result, next) in
27             let count = result[next] ?? 0
28             result[next] = count + 1
29         }
30     }
31 }

2652ms

 1 class Solution {
 2     func wordSubsets(_ A: [String], _ B: [String]) -> [String] {
 3 
 4         let bCounterArray = B
 5             .map {
 6                 $0.reduce(into: [:]) {
 7                     $0[String($1), default: 0] += 1
 8                 }
 9             }
10             .reduce(into: [:]) { (dict: inout [String: Int], word:  [String: Int]) in
11                 word.forEach {
12                     dict[$0] = (dict[$0] == nil ? $1 : max(dict[$0]!, $1))
13                 }
14         }
15 
16         let aCounterDictionary = A.reduce(into: [:]) {
17             $0[$1] = $1.reduce(into: [:]) {
18                 $0[String($1), default: 0] += 1
19             }
20         }
21 
22         var answer = [String]()
23 
24         for (word, aCounter) in aCounterDictionary {
25             var isUniversal = true
26             for (key, bval) in bCounterArray {
27                 guard let aval = aCounter[key], aval >= bval else {
28                     isUniversal = false
29                     break
30                 }
31             }
32 
33             if isUniversal {
34                 answer.append(word)
35             }
36         }
37 
38         return answer
39     }
40 }

2708ms

 1 class Solution {
 2     func wordSubsets(_ A: [String], _ B: [String]) -> [String] {
 3         var mb = [Character: Int]()
 4         for b1 in B {
 5             var d = [Character: Int]()
 6             for b2 in b1 {
 7                 d[b2, default: 0] += 1
 8             }
 9             for k in d.keys {
10                 mb[k, default: 0] = max(mb[k, default: 0], d[k]!)
11             }
12         }
13         //print(mb)
14         return A.filter { s in
15             var d = [Character: Int]()
16             for c in s {
17                 d[c, default: 0] += 1
18             }
19             //print(d)
20             for k in mb.keys {
21                 guard mb[k]! <= d[k, default: 0] else {
22                     return false
23                 }
24             }
25             return true
26         }
27     }
28 }

2712ms

 1 class Solution {
 2     func wordSubsets(_ A: [String], _ B: [String]) -> [String] {
 3         var result = [String]()
 4         var bMap = [Character : Int]()
 5         for bWord in B {
 6             var bMapTemp = [Character : Int]()
 7             for character in bWord {
 8                 if let value = bMapTemp[character] {
 9                     bMapTemp[character] = value + 1
10                 } else {
11                     bMapTemp[character] = 1
12                 }
13                 
14                 if let value1 = bMap[character], let value2 = bMapTemp[character]  {
15                     bMap[character] = max(value1, value2)
16                 } else {
17                     bMap[character] = bMapTemp[character]
18                 }
19     
20             }
21             
22         }
23         //print(bMap)
24         for aWord in A {
25             var aMap = [Character : Int]()
26             for character in aWord {
27                 if let value = aMap[character] {
28                     aMap[character] = value + 1
29                 } else {
30                     aMap[character] = 1
31                 }
32             }
33             
34             var shouldAdd = true
35             for (key, value) in bMap {
36                 //print("key \(key), value \(value), aMap \(aMap)")
37                 guard let aValue = aMap[key] else {
38                     shouldAdd = false
39                     break
40                 }
41                 if aValue < value {
42                     shouldAdd = false
43                     break
44                 }
45             }
46             if shouldAdd {
47                 result.append(aWord)
48             }
49         }
50         
51         return result
52     }
53 }

3208ms

 1 class Solution {
 2     
 3     // B = ["o", "oo", "oo", "ooo"]
 4     
 5     func wordSubsets(_ A: [String], _ B: [String]) -> [String] {
 6         let reducedSet: [Character: Int] = B.reduce(into: [:]) { res, next in
 7             for (k, v) in countCharacters(in: next) {
 8                 res[k] = max(v, res[k] ?? 0)
 9             }
10         }
11         return A.filter { self.isWord($0, universalFor: reducedSet) }
12     }
13     
14     func isWord(_ word: String, universalFor characters: [Character: Int]) -> Bool {
15         let characterCount = countCharacters(in: word)
16         return !characters.contains {
17             let count = characterCount[$0.key]
18             return count == nil || count! < $0.value
19         }
20     }
21     
22     func countCharacters(in word: String) -> [Character: Int] {
23         return word.reduce(into: [:]) { (result, next) in
24             let count = result[next] ?? 0
25             result[next] = count + 1
26         }
27     }
28 }

3600ms

 1 class Solution {
 2     
 3     // B = ["o", "oo", "oo", "ooo"]
 4     
 5     func wordSubsets(_ A: [String], _ B: [String]) -> [String] {
 6         let reducedSet: [Character: Int] = B.reduce(into: [:]) { res, next in
 7             for (k, v) in countCharacters(in: next) {
 8                 res[k] = max(v, res[k] ?? 0)
 9             }
10         }
11         return A.filter { self.isWord($0, universalFor: reducedSet) }
12     }
13     
14     func isWord(_ word: String, universalFor characters: [Character: Int]) -> Bool {
15         let characterCount = countCharacters(in: word)
16         return !characters.contains {
17             let count = characterCount[$0.key]
18             return count == nil || count! < $0.value
19         }
20     }
21     
22     func countCharacters(in word: String) -> [Character: Int] {
23         return word.reduce(into: [:]) { (result, next) in
24             let count = result[next] ?? 0
25             result[next] = count + 1
26         }
27     }
28 }

 

posted @ 2018-10-26 18:53  为敢技术  阅读(402)  评论(0)  编辑  收藏  举报