单词接龙 II （来自lintecode：https://www.lintcode.com/problem/word-ladder-ii/?_from=ladder&&fromId=1）
给出两个单词（start和end）和一个字典，找出所有从start到end的最短转换序列。
变换规则如下：
1.每次只能改变一个字母。
2.变换过程中的中间单词必须在字典中出现。
样例
给出数据如下：
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]
返回
[
["hit","hot","dot","dog","cog"],
["hit","hot","lot","log","cog"]
]
注意事项
1.所有单词具有相同的长度。
2.所有单词都只包含小写字母。
构造代码前的思路：
1.首先构造有向图，将start, end 以及dict里所有字符的可变化对象列在一个字典里。
2.计算每个字符距离终点的距离（BFS)
3.根据距离逐一添加进result (DFS)
操作实现：
import copy
class Solution:
graph = {}
used = set()
result = []
def findLadders(self, start, end, dict):
self.graph[start] = set()
self.graph[end] = set()
for word in dict:
self.graph[word] = set()#先建立起空的有向图
self.buildgraph(start, dict)
self.buildgraph(end, dict)
for word in dict:
self.buildgraph(word, dict)#开始建立有向图
dist = self.distance(end)#开始建立距离表
self.findresult(start, end, dist, [])#根据距离建立result
return self.result
def buildgraph(self, word, dict):#建立有向图
for char in range(len(word)):
left = word[:char]
right = word[char + 1:]
for mid in "abcdefghijklmnopwrstuvwxyz":
neword = left + mid + right
if neword != word and neword in dict:
self.graph[neword].add(word)#存在多个出度，所以是append而不是赋值等号
self.graph[word].add(neword)
def distance(self, word):
dist = {}
dist[word] = 0
self.used.add(word)#由于该有向图存在环，可以互相变化，为了避难重复使用导致覆盖距离
temp_list = [word]#为了BFS建立一个临时的表格
while temp_list:
word = temp_list.pop(0)#从最左开始取值，保证距离是从小到大
for neighbor in self.graph[word]:
if neighbor not in self.used:
self.used.add(neighbor)
temp_list.append(neighbor)
dist[neighbor] = dist[word] + 1
return dist
def findresult(self, word, end, dist, path):
path.append(word)
if word == end:
self.result.append(copy.deepcopy(path))
return
for neighbor in self.graph[word]:
if dist[neighbor] == dist[word] - 1: #如果距离相差为1，则录用
self.findresult(neighbor, end, dist, path)
path.pop()
class Solution:
graph = {}
used = set()
result = []
def findLadders(self, start, end, dict):
self.graph[start] = set()
self.graph[end] = set()
for word in dict:
self.graph[word] = set()#先建立起空的有向图
self.buildgraph(start, dict)
self.buildgraph(end, dict)
for word in dict:
self.buildgraph(word, dict)#开始建立有向图
dist = self.distance(end)#开始建立距离表
self.findresult(start, end, dist, [])#根据距离建立result
return self.result
def buildgraph(self, word, dict):#建立有向图
for char in range(len(word)):
left = word[:char]
right = word[char + 1:]
for mid in "abcdefghijklmnopwrstuvwxyz":
neword = left + mid + right
if neword != word and neword in dict:
self.graph[neword].add(word)#存在多个出度，所以是append而不是赋值等号
self.graph[word].add(neword)
def distance(self, word):
dist = {}
dist[word] = 0
self.used.add(word)#由于该有向图存在环，可以互相变化，为了避难重复使用导致覆盖距离
temp_list = [word]#为了BFS建立一个临时的表格
while temp_list:
word = temp_list.pop(0)#从最左开始取值，保证距离是从小到大
for neighbor in self.graph[word]:
if neighbor not in self.used:
self.used.add(neighbor)
temp_list.append(neighbor)
dist[neighbor] = dist[word] + 1
return dist
def findresult(self, word, end, dist, path):
path.append(word)
if word == end:
self.result.append(copy.deepcopy(path))
return
for neighbor in self.graph[word]:
if dist[neighbor] == dist[word] - 1: #如果距离相差为1，则录用
self.findresult(neighbor, end, dist, path)
path.pop()
