-
Notifications
You must be signed in to change notification settings - Fork 2
/
737. Sentence Similarity II.py
62 lines (54 loc) · 2.48 KB
/
737. Sentence Similarity II.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
# Given two sentences words1, words2 (each represented as an array of strings), and a list of similar word pairs pairs,
# determine if two sentences are similar.
# For example, words1 = ["great", "acting", "skills"] and words2 = ["fine", "drama", "talent"] are similar,
# if the similar word pairs are pairs = [["great", "good"], ["fine", "good"], ["acting","drama"], ["skills","talent"]].
# Note that the similarity relation is transitive.
# For example, if "great" and "good" are similar, and "fine" and "good" are similar, then "great" and "fine" are similar.
# Similarity is also symmetric.
# For example, "great" and "fine" being similar is the same as "fine" and "great" being similar. Also, a word is always similar with itself.
# For example, the sentences words1 = ["great"], words2 = ["great"], pairs = [] are similar,
# even though there are no specified similar word pairs.
# Finally, sentences can only be similar if they have the same number of words.
# So a sentence like words1 = ["great"] can never be similar to words2 = ["doubleplus","good"].
# Note:
# The length of words1 and words2 will not exceed 1000.
# The length of pairs will not exceed 2000.
# The length of each pairs[i] will be 2.
# The length of each words[i] and pairs[i][j] will be in the range [1, 20].
# Hints:
# Consider the graphs where each pair in "pairs" is an edge.
# Two words are similar if they are the same, or are in the same connected component of this graph.
class Solution(object):
def areSentencesSimilarTwo(self, words1, words2, pairs):
"""
:type words1: List[str]
:type words2: List[str]
:type pairs: List[List[str]]
:rtype: bool
"""
# 并查集
def find(word):
# 已经连通
if word in parents:
tmp = find(parents[word])
parents[word] = tmp
return tmp
# 尚未连通
return word
def union(parents, wd1, wd2):
p1 = find(wd1)
p2 = find(wd2)
if p1 != p2:
parents[p1] = p2
if not pairs:
return words1 == words2
if len(words1) != len(words2):
return False
parents = {}
# great good fine good
for pair in pairs:
union(parents, pair[0], pair[1])
for i in range(len(words1)):
if words1[i] != words2[i] and find(words1[i]) != find(words2[i]):
return False
return True