0126. Word Ladder I I

126. Word Ladder II #

题目 #

Given two words (beginWord and endWord), and a dictionary’s word list, find all shortest transformation sequence(s) 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 an empty list 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:
[
  ["hit","hot","dot","dog","cog"],
  ["hit","hot","lot","log","cog"]
]

Example 2:

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

Output: []

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 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
}


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