[Swift]LeetCode127.单词接龙|WordLadder

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Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time.
Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.
You may assume no duplicates in the word list.
You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output: 5

Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Example 2:

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output: 0

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

给定两个单词（beginWord 和 endWord）和一个字典，找到从 beginWord 到 endWord 的最短转换序列的长度。转换需遵循如下规则：

每次转换只能改变一个字母。
转换过程中的中间单词必须是字典中的单词。

说明:

如果不存在这样的转换序列，返回 0。
所有单词具有相同的长度。
所有单词只由小写字母组成。
字典中不存在重复的单词。
你可以假设 beginWord 和 endWord 是非空的，且二者不相同。

示例 1:

输入:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

输出: 5

解释: 一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog",
     返回它的长度 5。

示例 2:

输入:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

输出: 0

解释: endWord "cog" 不在字典中，所以无法进行转换。

304ms

 1 class Solution {
 2     func ladderLength(_ beginWord: String, _ endWord: String, _ wordList: [String]) -> Int {
 3         if !wordList.contains(endWord) 
 4         { 
 5             return 0 
 6         }
 7         
 8         let dict = Set(wordList)
 9         var beginSet = Set<String>()
10         var endSet = Set<String>()
11         var visitedSet = Set<String>()
12         var length = 1
13         var found = false
14         
15         beginSet.insert(beginWord)
16         endSet.insert(endWord)
17         
18         while !found && !beginSet.isEmpty && !endSet.isEmpty {
19             var nextSet = Set<String>()
20             //accelerating search speed by swap begin and end
21             if beginSet.count > endSet.count {
22                 swap(&beginSet, &endSet)
23             }
24             found = helper(beginSet, endSet, dict, &visitedSet, &nextSet)
25             beginSet = nextSet
26             length += 1
27         }
28         return found ? length : 0
29     }
30     
31     private func helper(_ beginSet: Set<String>, _ endSet: Set<String>, _ dict: Set<String>,
32                         _ visitedSet: inout Set<String>, _ resSet: inout Set<String>) -> Bool {
33         
34         let alphaArray = Array("abcdefghijklmnopqrstuvwxyz")
35         
36         for word in beginSet {
37             for i in 0 ..< word.count {
38                 var chars = Array(word)
39                 for j in 0 ..< alphaArray.count{
40                     chars[i] = alphaArray[j]
41                     let str = String(chars)
42                     if dict.contains(str) {
43                         if endSet.contains(str) 
44                         { 
45                             return true 
46                         }
47                         if !visitedSet.contains(str) 
48                         { 
49                             resSet.insert(str)
50                             visitedSet.insert(str) 
51                         }
52                     }
53                 }
54             }
55         }
56         return false
57     }
58 }

884ms

 1 class Solution {
 2     func ladderLength(_ beginWord: String, _ endWord: String, _ wordList: [String]) -> Int {
 3         // ensure the endWord is in the wordList
 4         if !wordList.contains(endWord) {
 5             return 0
 6         }
 7 
 8         // build a word map with char array representations
 9         // this saves an enormous amount of time
10         var wordMap: [String: [Character]] = [:]
11         for i in 0..<wordList.count {
12             // beginWord should never be in the list
13             if wordList[i] == beginWord {
14                 continue
15             }
16             wordMap[wordList[i]] = [Character](wordList[i].characters)
17         }
18         
19         // we work the problem from the begining forward and end backwards
20         // this allows us to deal with the smallest set of possibilities in 
21         // both "directions"
22         var beginMap:[String: [Character]] = [beginWord: [Character](beginWord.characters)]
23         var endMap:[String: [Character]] = [endWord: [Character](endWord.characters)]
24         
25         // burnedMap bridges what has been used in both beginMap or endMap
26         var burnedMap:[String: [Character]] = [:]
27         
28         // word len is stable throughout the iterations, pay for it only once
29         let wordLen = beginWord.count
30         
31         // mutations tracts how many time mutated on the road from beginWord to endWord
32         var mutations = 1
33 
34         // we continue while beginMap and endMap are non-empty
35         // if one empties, there is no path from beginWord to endWord
36         while !beginMap.isEmpty && !endMap.isEmpty {
37             // minimize the working set by swapping in the smaller
38             // set, we only care about the mutation count
39             if beginMap.count > endMap.count {
40                 let tempMap = beginMap
41                 beginMap = endMap
42                 endMap = tempMap
43             }
44 
45             // track the upcoming set of mutation candidates we're about to create
46             var newMap: [String: [Character]] = [:]
47 
48             // iterate throught he last set of mutation candidates
49             for candidateKV in beginMap {
50                 let candidateChars = candidateKV.value
51                 
52                 // walk through all the remaining (un-burned) valid words
53                 for wordKV in wordMap {
54                     let word = wordKV.key
55                     let wordChars = wordKV.value
56                     
57                     // diff the candidate and valid words looking
58                     // fir diffs of 1 character
59                     var diffs = 0
60                     
61                     for i in 0..<wordLen {
62                         if candidateChars[i] != wordChars[i] {
63                             diffs += 1
64                             if diffs > 1 {
65                                 break
66                             }
67                         }
68                     }
69                     
70                     // anything other than a diff of 1 is not applicable
71                     // in this loop, we can only consider "word" if the
72                     // diff is exactly 1
73                     if diffs == 1 {
74                         // we have a valid 1 character diff, use it
75                         
76                         if endMap[word] != nil {
77                             // base case reached
78                             return mutations + 1
79                         }
80                         
81                         if burnedMap[word] == nil {
82                             newMap[word] = wordChars
83                             burnedMap[word] = wordChars
84                         }
85                     }
86                 }
87             }
88             
89             // we already evaluated what was in beginMap, we can 
90             // lose those and use the newMap evaluation set 
91             beginMap = newMap
92             
93             // we have completed an additional mutation step
94             mutations += 1
95         }
96         
97         return 0
98     }
99 }

1520ms

  1 class Solution {
  2     func ladderLength(_ beginWord: String, _ endWord: String, _ wordList: [String]) -> Int {
  3         //Set类型替换数组类型，加速查找
  4         var wordSet:Set<String> = Set<String>(wordList)
  5         //构建队列，实现广度优先遍历
  6         var q:Queue<String> = Queue<String>()
  7         //加入源顶点
  8         q.enQueue(beginWord)
  9         var res:Int = 0
 10         while (!q.isEmpty())
 11         {
 12             for k in (1...q.count).reversed()
 13             {
 14                 var word:String = q.deQueue()!
 15                 if word == endWord {return res + 1}
 16                 for i in 0..<word.count
 17                 {
 18                     var newWord:String = word
 19                     for ch in 97...122
 20                     {
 21                         newWord[i] = ch.ASCII 
 22                         if wordSet.contains(newWord) && newWord != word
 23                         {
 24                             q.enQueue(newWord)
 25                             wordSet.remove(newWord)
 26                         }
 27                     }
 28                 }
 29             }
 30             res += 1
 31         }
 32          return 0
 33     }
 34 }
 35 
 36 //String扩展方法
 37 extension String {
 38     func toCharArray() -> [Character]
 39     {
 40         var arr:[Character] = [Character]()
 41         for char in self.characters
 42         {
 43             arr.append(char)
 44         }
 45         return arr
 46     }
 47     
 48     //subscript函数可以检索数组中的值
 49     //直接按照索引方式截取指定索引的字符
 50     subscript (_ i: Int) -> Character {
 51         //读取字符
 52         get {return self[index(startIndex, offsetBy: i)]}
 53         
 54         //修改字符
 55         set
 56         {
 57             var str:String = self
 58             var index = str.index(startIndex, offsetBy: i)
 59             str.remove(at: index)
 60             str.insert(newValue, at: index)
 61             self = str
 62         }
 63     }
 64 }
 65 
 66 //Int扩展方法  
 67 extension Int
 68 {
 69     //属性：ASCII值(定义大写为字符值)
 70     var ASCII:Character 
 71     {
 72         get {return Character(UnicodeScalar(self)!)}
 73     }
 74 }
 75 
 76 public struct Queue<T> {
 77     
 78     // 泛型数组：用于存储数据元素
 79     fileprivate var queue: [T] 
 80 
 81     // 返回队列中元素的个数
 82     public var count: Int {
 83         return queue.count
 84     }
 85     
 86     // 构造函数：创建一个空的队列
 87     public init() {
 88         queue = [T]()
 89     }
 90     
 91     //通过既定数组构造队列
 92     init(_ arr:[T]){
 93         queue = arr
 94     }
 95     
 96     // 检查队列是否为空
 97     // - returns: 如果队列为空，则返回true，否则返回false
 98     public func isEmpty() -> Bool {
 99         return queue.isEmpty
100     }
101     
102     // 入队列操作：将元素添加到队列的末尾
103     public mutating func enQueue(_ element: T) {
104         queue.append(element)
105     }
106     
107     // 出队列操作：删除并返回队列中的第一个元素
108     public mutating func deQueue() -> T? {
109         return queue.removeFirst()
110     }
111 }

1692ms

 1 class Solution {
 2     func ladderLength(_ beginWord: String, _ endWord: String, _ wordList: [String]) -> Int {
 3         
 4         var wordSet = Set(wordList)
 5         var level = 0
 6         var currentLevelWords = [beginWord]
 7         
 8         while !currentLevelWords.isEmpty {
 9             level += 1
10             
11             var nextLevelWords: [String] = []
12             
13             for word in currentLevelWords {
14                 if word == endWord {
15                     return level
16                 }
17                 
18                 nextLevelWords += findRelatedWords(word: word, wordSet: &wordSet)
19             }
20             
21             currentLevelWords = nextLevelWords
22         }
23         
24         return 0
25     }
26     
27     private let alphabet = Array("abcdefghijklmnopqrstuvwxyz")
28     
29     func findRelatedWords(word: String, wordSet: inout Set<String>) -> [String] {
30         var relatedWords: [String] = []
31         
32         var word = word
33         var wordIndex = word.startIndex
34         while wordIndex < word.endIndex {
35             
36             let originalCharacter = word[wordIndex]
37             
38             for character in alphabet {
39                 word.remove(at: wordIndex)
40                 word.insert(character, at: wordIndex)
41                 
42                 if wordSet.contains(word) {
43                     wordSet.remove(word)
44                     
45                     relatedWords.append(word)
46                 }
47             }
48             
49             word.remove(at: wordIndex)
50             word.insert(originalCharacter, at: wordIndex)
51             
52             wordIndex = word.index(after: wordIndex)
53         }
54         
55         return relatedWords
56     }
57 }

1788ms

 1 class Solution {
 2     func ladderLength(_ beginWord: String, _ endWord: String, _ wordList: [String]) -> Int {
 3         
 4         var wordSet = Set(wordList)
 5         var currentLevelWords = [beginWord]
 6         var level = 0
 7         
 8         while !currentLevelWords.isEmpty {
 9             level += 1
10             
11             var nextLevelWords: [String] = []
12             
13             for word in currentLevelWords {
14                 
15                 if word == endWord {
16                     return level
17                 }
18                 
19                 nextLevelWords += findCloseWords(word: word, wordSet: &wordSet)
20             }
21             
22             currentLevelWords = nextLevelWords
23         }
24         
25         return 0
26     }
27     
28     let alphabet = Array("abcdefghijklmnopqrstuvwxyz")
29     
30     func findCloseWords(word: String, wordSet: inout Set<String>) -> [String] {
31         
32         var closeWords: [String] = []
33         
34         var word = word
35         var wordIndex = word.startIndex
36         
37         while wordIndex < word.endIndex {
38             
39             let originalCharacter = word[wordIndex]
40             
41             for character in alphabet {
42                 word.remove(at: wordIndex)
43                 word.insert(character, at: wordIndex)
44                 
45                 if wordSet.contains(word) {
46                     closeWords.append(word)
47                     wordSet.remove(word)
48                 }
49             }
50             
51             word.remove(at: wordIndex)
52             word.insert(originalCharacter, at: wordIndex)
53             
54             wordIndex = word.index(after: wordIndex)
55         }
56         
57         return closeWords
58     }
59 }

2284ms

 1 class Solution {
 2     func ladderLength(_ beginWord: String, _ endWord: String, _ wordList: [String]) -> Int {
 3         var queue = [beginWord]
 4         var wordList = Set(wordList)
 5         var length = 1
 6 
 7         while !queue.isEmpty && !wordList.isEmpty {
 8             let count = queue.count
 9             for _ in 0..<count {
10                 let word = queue.removeFirst()
11 
12                 let validTransforms = validTransformations(word, wordList)
13                 for nextWord in validTransforms {
14                     if nextWord == endWord {
15                         return length + 1
16                     }
17                     wordList.remove(nextWord)
18                     queue.append(nextWord)
19                 }
20             }
21 
22             length += 1
23         }
24 
25         return 0
26     }
27     
28     func validTransformations(_ word: String, _ wordList: Set<String>) -> [String] {
29         var result = [String]()
30         let alpahabetArray = Array("abcdefghijklmnopqrstuvwxyz".characters)
31         for i in 0 ..< word.characters.count {
32             var chars = Array(word.characters)
33              
                       
                    
                     
                      



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