Hard
A transformation sequence from word `beginWord` to word `endWord` using a dictionary `wordList` is a sequence of words `beginWord -> s1 -> s2 -> ... -> sk` such that:
• Every adjacent pair of words differs by a single letter.
• Every `si` for `1 <= i <= k` is in `wordList`. Note that `beginWord` does not need to be in `wordList`.
• `sk == endWord`
Given two words, `beginWord` and `endWord`, and a dictionary `wordList`, return the number of words in the shortest transformation sequence from `beginWord` to `endWord`, or `0` if no such sequence exists.
Example 1:
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output:
5
Explanation:
One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.
Example 2:
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output:
0
Explanation:
The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.
Constraints:
• `1 <= beginWord.length <= 10`
• `endWord.length == beginWord.length`
• `1 <= wordList.length <= 5000`
• `wordList[i].length == beginWord.length`
• `beginWord`, `endWord`, and `wordList[i]` consist of lowercase English letters.
• `beginWord != endWord`
• All the words in `wordList` are unique.

### 解題

Runtime: 150 ms, faster than 62.69%
Memory Usage: 7.8 MB, less than 46.71%
type Pair struct {
word string
distance int
}
func ladderLength(beginWord string, endWord string, wordList []string) int {
visited := make(map[string]bool)
wordMap := make(map[string]bool)
queue := []Pair{ Pair{beginWord, 0} }
for _, word := range wordList {
wordMap[word] = true
}
if !wordMap[endWord] { return 0 }
visited[beginWord] = true
for len(queue) != 0 {
cur := queue[0]
queue = queue[1:]
if cur.word == endWord { return cur.distance + 1 }
bytes := []byte(cur.word)
for i := 0; i < len(bytes); i++ {
ch := bytes[i]
for j := 0; j < 26; j++ {
rep := j + 'a'
bytes[i] = byte(rep)
tmp := string(bytes)
if _, ok := wordMap[tmp]; ok && !visited[tmp] {
queue = append(queue, Pair{tmp, cur.distance + 1 })
visited[tmp] = true
}
}
bytes[i] = ch
}
}
return 0
}