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 #

给定两个单词(beginWord 和 endWord)和一个字典,找到从 beginWord 到 endWord 的最短转换序列的长度。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 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
}


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