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描述
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.
Note:
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.
解析
根据题意，就是给出了一个 beginWord ，经过在 wordList 中一系列的查找，最后能得到 endWord 的单词序列长度，这里的查找要求每次只能变化一个字母。其实这个题就是类似于走迷宫，只不过每次变化的可能从 4 个变成了 26 个。这里我用集合 visited 来存储已经用过的单词，然后利用栈变量 queue 保存查找到的有效的单词，再对栈中的每个单词进行进一步深入的变化和查找，将可能的单词再追加入栈中，直到找到 endWord 为止，并返回这个寻找过程中产生的变化序列的距离。
不知道为何会超时，尴了个尬。这种方法可能还是存在缺陷吧，也可能是题目的限制比较大，可以换一种思路继续进行解题。
解答
class Solution(object):
def ladderLength(self, beginWord, endWord, wordList):
"""
:type beginWord: str
:type endWord: str
:type wordList: List[str]
:rtype: int
"""
queue = [(beginWord, 1)]
visited = set()
while queue:
word, dist = queue.pop(0)
if word == endWord:
return dist
for i in range(len(word)):
for j in 'abcdefghijklmnopqrstuvwxyz':
tmp = word[:i] + j + word[i+1:]
if tmp not in visited and tmp in wordList:
queue.append((tmp, dist+1))
visited.add(tmp)
return 0
运行结果
Time Limit Exceeded
解析
换一种思路，很多时间都浪费在了制造出来的 tmp 判断是否已经用过并且是否存在于 wordList 种，我们可以缩减这一部分的时间，如果这个制造出来的单词出现在 wordList ，说明就会一定用到它，所以不需要记录在 visited ，直接将其从 wordList 删除即可，这样可以减少布尔语句的判断时间，并且随着查找的进行不断删减单词， wordList 残留会越来越少，也会同时减少查找的时间。这次没有超时，但是耗时还是很多。
解答
class Solution(object):
def ladderLength(self, beginWord, endWord, wordList):
"""
:type beginWord: str
:type endWord: str
:type wordList: List[str]
:rtype: int
"""
wordList = set(wordList)
queue = [(beginWord, 1)]
for word, dist in queue:
if word == endWord:
return dist
for i in range(len(word)):
for char in 'abcdefghijklmnopqrstuvwxyz':
tmp = word[:i] + char + word[i + 1:]
if tmp in wordList:
queue.append([tmp, dist + 1])
wordList.remove(tmp)
return 0
运行结果
Runtime: 513 ms, faster than 25.82% of Python online submissions for Word Ladder.
Memory Usage: 14.6 MB, less than 62.38% of Python online submissions for Word Ladder.
原题链接：leetcode.com/problems/wo…
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