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 }
原文地址：https://www.cnblogs.com/strengthen/p/9958336.html
时间： 2024-10-27 19:07:02