Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord, such that: Only one letter can be changed at a time Each transformed word must exist in the word list. Note that beginWord is not a transformed word. For example, Given: beginWord = "hit" endWord = "cog" wordList = ["hot","dot","dog","lot","log","cog"] Return [ ["hit","hot","dot","dog","cog"], ["hit","hot","lot","log","cog"] ] 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. UPDATE (2017/1/20): The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.
from collections import deque
class Solution(object):
def findLadders(self, beginWord, endWord, wordlist):
"""
:type beginWord: str
:type endWord: str
:type wordlist: Set[str]
:rtype: List[List[int]]
"""
def getNbrs(src, dest, wordList):
res = []
for c in string.ascii_lowercase:
for i in xrange(0, len(src)):
newWord = src[:i] + c + src[i+1:]
if newWord == src:
continue
if newWord in wordList or newWord == dest:
yield newWord
def bfs(beginWord, endWord, wordList):
distance = {beginWord: 0}
queue = deque([beginWord])
length = 0
while queue:
length += 1
for k in xrange(0, len(queue)):
top = queue.popleft()
for nbr in getNbrs(top, endWord, wordList):
if nbr not in distance:
distance[nbr] = distance[top] + 1
queue.append(nbr)
return distance
def dfs(beginWord, endWord, wordList, path, res, distance):
if beginWord == endWord:
res.append(path + [])
return
for nbr in getNbrs(beginWord, endWord, wordList):
if distance.get(nbr, -2) + 1 == distance[beginWord]:
path.append(nbr)
dfs(nbr, endWord, wordList, path, res, distance)
path.pop()
res = []
distance = bfs(endWord, beginWord, wordlist)
dfs(beginWord, endWord, wordlist, [beginWord], res, distance)
return res
