0127. Word Ladder

# 127. Word Ladder#

## 题目 #

Given two words (beginWord and endWord), and a dictionary’s word list, find the length of shortest transformation sequence 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 0 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: 5

Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.
``````

Example 2:

``````Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output: 0

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
``````

## 题目大意 #

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。

• 如果不存在这样的转换序列，返回 0。
• 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 #

• 这一题要求输出从 `beginWord` 变换到 `endWord` 最短变换次数。可以用 BFS，从 `beginWord` 开始变换，把该单词的每个字母都用 `'a'~'z'` 变换一次，生成的数组到 `wordList` 中查找，这里用 Map 来记录查找。找得到就入队列，找不到就输出 0 。入队以后按照 BFS 的算法依次遍历完，当所有单词都 `len(queue)<=0` 出队以后，整个程序结束。
• 这一题题目中虽然说了要求找到一条最短的路径，但是实际上最短的路径的寻找方法已经告诉你了：
1. 每次只变换一个字母
2. 每次变换都必须在 `wordList`
所以不需要单独考虑何种方式是最短的。

## 代码 #

``````
package leetcode

func ladderLength(beginWord string, endWord string, wordList []string) int {
wordMap, que, depth := getWordMap(wordList, beginWord), []string{beginWord}, 0
for len(que) > 0 {
depth++
qlen := len(que)
for i := 0; i < qlen; i++ {
word := que[0]
que = que[1:]
candidates := getCandidates(word)
for _, candidate := range candidates {
if _, ok := wordMap[candidate]; ok {
if candidate == endWord {
return depth + 1
}
delete(wordMap, candidate)
que = append(que, candidate)
}
}
}
}
return 0
}

func getWordMap(wordList []string, beginWord string) map[string]int {
wordMap := make(map[string]int)
for i, word := range wordList {
if _, ok := wordMap[word]; !ok {
if word != beginWord {
wordMap[word] = i
}
}
}
return wordMap
}

func getCandidates(word string) []string {
var res []string
for i := 0; i < 26; i++ {
for j := 0; j < len(word); j++ {
if word[j] != byte(int('a')+i) {
res = append(res, word[:j]+string(int('a')+i)+word[j+1:])
}
}
}
return res
}

``````

Apr 8, 2023
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