[Swift]LeetCode126. 单词接龙 II | Word Ladder II

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/9958336.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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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

1. Only one letter can be changed at a time
2. 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：

1. 每次转换只能改变一个字母。
2. 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

class Solution {
    let alphabet = Array("abcdefghijklmnopqrstuvwxyz")

    func findLadders(_ beginWord: String, _ endWord: String, _ wordList: [String]) -> [[String]] {
        var dictionary = Set(wordList)
        var result = [[String]]()
        var distance = [String: Int]()
        var neighbors = [String: [String]]()

        dictionary.insert(beginWord)

        // Get distances and neighbors for each word
        bfs(beginWord, endWord, &dictionary, &distance, &neighbors)

        // Get results of word ledders
        var temp = [String]()
        dfs(beginWord, endWord, &dictionary, &distance, &neighbors, &result, &temp)

        return result
    }

    private func bfs(_ beginWord: String, _ endWord: String, _ dictionary: inout Set<String>, _ distance: inout [String: Int], _ neighbors: inout [String: [String]]) {
        for word in dictionary {
            neighbors[word] = [String]()
        }

        var queue = [String]()
        queue.append(beginWord)
        distance[beginWord] = 0

        while !queue.isEmpty {
            var newQueue = [String]()
            var foundEnd = false
            for word in queue {
                let wordDistance = distance[word]!
                let wordNeighbors = getNeighbors(&dictionary, word)
                for neighbor in wordNeighbors {
                    neighbors[word]!.append(neighbor)
                    if distance[neighbor] == nil {
                        distance[neighbor] = wordDistance + 1
                        if neighbor == endWord {
                            foundEnd = true
                        } else {
                            newQueue.append(neighbor)
                        }
                    }
                }
            }
            if foundEnd {
                break
            }
            queue = newQueue
        }
    }

    private func getNeighbors(_ dictionary: inout Set<String>, _ word: String) -> [String] {
        var wordChars = Array(word)
        var result = [String]()
        for i in 0..<word.count {
            let oldChar = wordChars[i]
            for letter in alphabet {
                wordChars[i] = letter
                let newWord = String(wordChars)
                if dictionary.contains(newWord) {
                    result.append(newWord)
                }
            }
            wordChars[i] = oldChar
        }
        return result
    }

    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]) {
        temp.append(beginWord)
        if beginWord == endWord {
            result.append(temp)
        } else {
            let wordDistance = distance[beginWord]!
            for neighbor in neighbors[beginWord]! {
                if distance[neighbor]! == wordDistance + 1 {
                    dfs(neighbor, endWord, &dictionary, &distance, &neighbors, &result, &temp)
                }
            }
        }
        temp.removeLast()
    }
}

---

4284ms

class Solution {
    func findLadders(_ beginWord: String, _ endWord: String, _ wordList: [String]) -> [[String]] {
        var res:[[String]] = [[String]]()
         //加速查找
        var dict:Set<String> = Set<String>(wordList)
        //构建队列
        var p:[String] = [String]()
        //加入源顶点
        p.append(beginWord)
        var paths:[[String]] = [[String]]()
        paths.append(p)
        var level:Int = 1, minLevel:Int = Int.max
        var words:Set<String> = Set<String>()
        while (paths.count != 0)
        {
            var t = paths.removeFirst()
            if t.count > level
            {
                for w in words{dict.remove(w)}
                words = Set<String>()
                level = t.count
                if level > minLevel {break}
            }
            var last:String = t.last!
            for i in 0..<last.count
            {
                var newLast:String = last
                for ch in 97...122
                {
                    newLast[i] = ch.ASCII
                    if !dict.contains(newLast){continue}
                    words.insert(newLast)
                    var nextPath = t
                    nextPath.append(newLast)
                    if newLast == endWord
                    {
                        res.append(nextPath)
                        minLevel = level
                    }
                    else
                    {
                        paths.append(nextPath)
                    }
                }
            }
        }
        return res
    }
}

//String扩展方法
extension String {
    func toCharArray() -> [Character]
    {
        var arr:[Character] = [Character]()
        for char in self.characters
        {
            arr.append(char)
        }
        return arr
    }
    
    //subscript函数可以检索数组中的值
    //直接按照索引方式截取指定索引的字符
    subscript (_ i: Int) -> Character {
        //读取字符
        get {return self[index(startIndex, offsetBy: i)]}
        
        //修改字符
        set
        {
            var str:String = self
            var index = str.index(startIndex, offsetBy: i)
            str.remove(at: index)
            str.insert(newValue, at: index)
            self = str
        }
    }
}

//Int扩展方法  
extension Int
{
    //属性：ASCII值(定义大写为字符值)
    var ASCII:Character 
    {
        get {return Character(UnicodeScalar(self)!)}
    }
}