127. Word Ladder ⭐

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
這題和 433. 非常相似～ 下面是我參考了討論區的 struct 架構後的解法。
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
}
​
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