📑 题目：126. 单词接龙 II

🚀 本题 LeetCode 传送门

题目大意

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。

解题思路

• 这一题是第 127 题的加强版，除了找到路径的长度，还进一步要求输出所有路径。解题思路同第 127 题一样，也是用 BFS 遍历。
• 当前做法不是最优解，是否可以考虑双端 BFS 优化，或者迪杰斯塔拉算法？

代码

package leetcode
func findLadders(beginWord string, endWord string, wordList []string) [][]string {
 result, wordMap := make([][]string, 0), make(map[string]bool)
 for _, w := range wordList {
  wordMap[w] = true
 }
 if !wordMap[endWord] {
  return result
 }
 // create a queue, track the path
 queue := make([][]string, 0)
 queue = append(queue, []string{beginWord})
 // queueLen is used to track how many slices in queue are in the same level
 // if found a result, I still need to finish checking current level cause I need to return all possible paths
 queueLen := 1
 // use to track strings that this level has visited
 // when queueLen == 0, remove levelMap keys in wordMap
 levelMap := make(map[string]bool)
 for len(queue) > 0 {
  path := queue[0]
  queue = queue[1:]
  lastWord := path[len(path)-1]
  for i := 0; i < len(lastWord); i++ {
   for c := 'a'; c <= 'z'; c++ {
    nextWord := lastWord[:i] + string(c) + lastWord[i+1:]
    if nextWord == endWord {
     path = append(path, endWord)
     result = append(result, path)
     continue
    }
    if wordMap[nextWord] {
     // different from word ladder, don't remove the word from wordMap immediately
     // same level could reuse the key.
     // delete from wordMap only when currently level is done.
     levelMap[nextWord] = true
     newPath := make([]string, len(path))
     copy(newPath, path)
     newPath = append(newPath, nextWord)
     queue = append(queue, newPath)
    }
   }
  }
  queueLen--
  // if queueLen is 0, means finish traversing current level. if result is not empty, return result
  if queueLen == 0 {
   if len(result) > 0 {
    return result
   }
   for k := range levelMap {
    delete(wordMap, k)
   }
   // clear levelMap
   levelMap = make(map[string]bool)
   queueLen = len(queue)
  }
 }
 return result
}