本文介绍了如何计算"最短距离和QUOT;两个词之间?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
最近我有一次采访,我被要求写一个算法来找出1号更改的最低数量从一个特定的词,以获得给定的话,即总分类> Cot-> Cog->狗
Recently I had an interview and I was asked to write a algorithm to find the minimum number of 1 letter changes to get from a particular of word to a given word , i.e. Cat->Cot->Cog->Dog
我不希望这个问题的解决方案,只是引导我通过我怎么能在这个算法中使用BFS?
I dont want the solution of the problem just guide me through How I can use BFS in this algorithm ?
推荐答案
根据这个拼字游戏列表,猫与狗之间的最短路径是:['猫','COT,COG','狗']
according to this scrabble list, the shortest path between cat and dog is:['CAT', 'COT', 'COG', 'DOG']
from urllib import urlopen
def get_words():
try:
html = open('three_letter_words.txt').read()
except IOError:
html = urlopen('http://www.yak.net/kablooey/scrabble/3letterwords.html').read()
with open('three_letter_words.txt', 'w') as f:
f.write(html)
b = html.find('<PRE>') #ignore the html before the <pre>
while True:
a = html.find("<B>", b) + 3
b = html.find("</B>", a)
word = html[a: b]
if word == "ZZZ":
break
assert(len(word) == 3)
yield word
words = list(get_words())
def get_template(word):
c1, c2, c3 = word[0], word[1], word[2]
t1 = 1, c1, c2
t2 = 2, c1, c3
t3 = 3, c2, c3
return t1, t2, t3
d = {}
for word in words:
template = get_template(word)
for ti in template:
d[ti] = d.get(ti, []) + [word] #add the word to the set of words with that template
for ti in get_template('COG'):
print d[ti]
#['COB', 'COD', 'COG', 'COL', 'CON', 'COO', 'COO', 'COP', 'COR', 'COS', 'COT', 'COW', 'COX', 'COY', 'COZ']
#['CIG', 'COG']
# ['BOG', 'COG', 'DOG', 'FOG', 'HOG', 'JOG', 'LOG', 'MOG', 'NOG', 'TOG', 'WOG']
import networkx
G = networkx.Graph()
for word_list in d.values():
for word1 in word_list:
for word2 in word_list:
if word1 != word2:
G.add_edge(word1, word2)
print G['COG']
#{'COP': {}, 'COS': {}, 'COR': {}, 'CIG': {}, 'COT': {}, 'COW': {}, 'COY': {}, 'COX': {}, 'COZ': {}, 'DOG': {}, 'CON': {}, 'COB': {}, 'COD': {}, 'COL': {}, 'COO': {}, 'LOG': {}, 'TOG': {}, 'JOG': {}, 'BOG': {}, 'HOG': {}, 'FOG': {}, 'WOG': {}, 'NOG': {}, 'MOG': {}}
print networkx.shortest_path(G, 'CAT', 'DOG')
['CAT', 'OCA', 'DOC', 'DOG']
作为奖励,我们可以得到最远的:
As a bonus we can get the farthest:
print max(networkx.all_pairs_shortest_path(G, 'CAT')['CAT'].values(), key=len)
#['CAT', 'CAP', 'YAP', 'YUP', 'YUK']
这篇关于如何计算&QUOT;最短距离和QUOT;两个词之间?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!