LeetCode - #126 单词接龙 II

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前言

本题由于没有合适答案为以往遗留问题，最近有时间将以往遗留问题一一完善。

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LeetCode 算法到目前我们已经更新到 125 期，我们会保持更新时间和进度（周一、周三、周五早上 9:00 发布），每期的内容不多，我们希望大家可以在上班路上阅读，长久积累会有很大提升。

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难度水平：困难

1. 描述

按字典 wordList 完成从单词 beginWord 到单词 endWord 转化，一个表示此过程的 转换序列 是形式上像 beginWord -> s1 -> s2 -> ... -> sk 这样的单词序列，并满足：

• 每对相邻的单词之间仅有单个字母不同。
• 转换过程中的每个单词 si（1 <= i <= k）必须是字典 wordList 中的单词。注意，beginWord 不必是字典 wordList 中的单词。
• sk == endWord

给你两个单词 beginWord 和 endWord ，以及一个字典 wordList 。请你找出并返回所有从 beginWord 到 endWord 的 最短转换序列 ，如果不存在这样的转换序列，返回一个空列表。每个序列都应该以单词列表 [beginWord, s1, s2, ..., sk] 的形式返回。

2. 示例

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"]]
解释：存在 2 种最短的转换序列：
"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 wordList 中，所以不存在符合要求的转换序列。

提示：

• 1 <= beginWord.length <= 5
• endWord.length == beginWord.length
• 1 <= wordList.length <= 500
• wordList[i].length == beginWord.length
• beginWord、endWord 和 wordList[i] 由小写英文字母组成
• beginWord != endWord
• wordList 中的所有单词 互不相同

3. 答案

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()
   }
}

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