Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) 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 an empty list 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: [ ["hit","hot","dot","dog","cog"], ["hit","hot","lot","log","cog"] ]
Example 2:
Input: beginWord = "hit" endWord = "cog" wordList = ["hot","dot","dog","lot","log"] Output: [] 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 wordList，找出所有从 beginWord 到 endWord 的最短转换序列。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，返回一个空列表。
- 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: [ ["hit","hot","dot","dog","cog"], ["hit","hot","lot","log","cog"] ]
示例 2:
输入: beginWord = "hit" endWord = "cog" wordList = ["hot","dot","dog","lot","log"] 输出: [] 解释: endWord "cog" 不在字典中，所以不存在符合要求的转换序列。
3640ms
1 class Solution { 2 let alphabet = Array("abcdefghijklmnopqrstuvwxyz") 3 4 func findLadders(_ beginWord: String, _ endWord: String, _ wordList: [String]) -> [[String]] { 5 var dictionary = Set(wordList) 6 var result = [[String]]() 7 var distance = [String: Int]() 8 var neighbors = [String: [String]]() 9 10 dictionary.insert(beginWord) 11 12 // Get distances and neighbors for each word 13 bfs(beginWord, endWord, &dictionary, &distance, &neighbors) 14 15 // Get results of word ledders 16 var temp = [String]() 17 dfs(beginWord, endWord, &dictionary, &distance, &neighbors, &result, &temp) 18 19 return result 20 } 21 22 private func bfs(_ beginWord: String, _ endWord: String, _ dictionary: inout Set<String>, _ distance: inout [String: Int], _ neighbors: inout [String: [String]]) { 23 for word in dictionary { 24 neighbors[word] = [String]() 25 } 26 27 var queue = [String]() 28 queue.append(beginWord) 29 distance[beginWord] = 0 30 31 while !queue.isEmpty { 32 var newQueue = [String]() 33 var foundEnd = false 34 for word in queue { 35 let wordDistance = distance[word]! 36 let wordNeighbors = getNeighbors(&dictionary, word) 37 for neighbor in wordNeighbors { 38 neighbors[word]!.append(neighbor) 39 if distance[neighbor] == nil { 40 distance[neighbor] = wordDistance + 1 41 if neighbor == endWord { 42 foundEnd = true 43 } else { 44 newQueue.append(neighbor) 45 } 46 } 47 } 48 } 49 if foundEnd { 50 break 51 } 52 queue = newQueue 53 } 54 } 55 56 private func getNeighbors(_ dictionary: inout Set<String>, _ word: String) -> [String] { 57 var wordChars = Array(word) 58 var result = [String]() 59 for i in 0..<word.count { 60 let oldChar = wordChars[i] 61 for letter in alphabet { 62 wordChars[i] = letter 63 let newWord = String(wordChars) 64 if dictionary.contains(newWord) { 65 result.append(newWord) 66 } 67 } 68 wordChars[i] = oldChar 69 } 70 return result 71 } 72 73 private func dfs(_ beginWord: String, _ endWord: String, _ dictionary: inout Set<String>, _ distance: inout [String: Int], _ neighbors: inout [String: [String]], _ result: inout [[String]], _ temp: inout [String]) { 74 temp.append(beginWord) 75 if beginWord == endWord { 76 result.append(temp) 77 } else { 78 let wordDistance = distance[beginWord]! 79 for neighbor in neighbors[beginWord]! { 80 if distance[neighbor]! == wordDistance + 1 { 81 dfs(neighbor, endWord, &dictionary, &distance, &neighbors, &result, &temp) 82 } 83 } 84 } 85 temp.removeLast() 86 } 87 }
4284ms
1 class Solution { 2 func findLadders(_ beginWord: String, _ endWord: String, _ wordList: [String]) -> [[String]] { 3 var res:[[String]] = [[String]]() 4 //加速查找 5 var dict:Set<String> = Set<String>(wordList) 6 //构建队列 7 var p:[String] = [String]() 8 //加入源顶点 9 p.append(beginWord) 10 var paths:[[String]] = [[String]]() 11 paths.append(p) 12 var level:Int = 1, minLevel:Int = Int.max 13 var words:Set<String> = Set<String>() 14 while (paths.count != 0) 15 { 16 var t = paths.removeFirst() 17 if t.count > level 18 { 19 for w in words{dict.remove(w)} 20 words = Set<String>() 21 level = t.count 22 if level > minLevel {break} 23 } 24 var last:String = t.last! 25 for i in 0..<last.count 26 { 27 var newLast:String = last 28 for ch in 97...122 29 { 30 newLast[i] = ch.ASCII 31 if !dict.contains(newLast){continue} 32 words.insert(newLast) 33 var nextPath = t 34 nextPath.append(newLast) 35 if newLast == endWord 36 { 37 res.append(nextPath) 38 minLevel = level 39 } 40 else 41 { 42 paths.append(nextPath) 43 } 44 } 45 } 46 } 47 return res 48 } 49 } 50 51 //String扩展方法 52 extension String { 53 func toCharArray() -> [Character] 54 { 55 var arr:[Character] = [Character]() 56 for char in self.characters 57 { 58 arr.append(char) 59 } 60 return arr 61 } 62 63 //subscript函数可以检索数组中的值 64 //直接按照索引方式截取指定索引的字符 65 subscript (_ i: Int) -> Character { 66 //读取字符 67 get {return self[index(startIndex, offsetBy: i)]} 68 69 //修改字符 70 set 71 { 72 var str:String = self 73 var index = str.index(startIndex, offsetBy: i) 74 str.remove(at: index) 75 str.insert(newValue, at: index) 76 self = str 77 } 78 } 79 } 80 81 //Int扩展方法 82 extension Int 83 { 84 //属性：ASCII值(定义大写为字符值) 85 var ASCII:Character 86 { 87 get {return Character(UnicodeScalar(self)!)} 88 } 89 }