Problem:
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, “great acting skills” and “fine drama talent” are similar, if the similar word pairs are pairs = [["great", "fine"], ["acting","drama"], ["skills","talent"]].
Note that the similarity relation is not transitive. For example, if “great” and “fine” are similar, and “fine” and “good” are similar, “great” and “good” are not necessarily similar.
However, similarity is 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
words1andwords2will not exceed1000. - The length of
pairswill not exceed2000. - The length of each
pairs[i]will be2. - The length of each
words[i]andpairs[i][j]will be in the range[1, 20].
题目大意：
给你两个句子（由单词数组表示）和一些近义词对，问你这两个句子是否相似，即每组相对应的单词都要相似。
注意相似性不能传递，比如给只你”great”和”fine”相似、”fine”和”good”相似，这种情况下”great”和”good”不相似。
Idea:
Use hashtable to store mapping from word to its similar words.
Solution: HashTable
Time complexity: O(|pairs| + |words1|)
Space complexity: O(|pairs|)
C++
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// Author: Huahua // Runtime: 9 ms class Solution { public: bool areSentencesSimilar(vector<string>& words1, vector<string>& words2, vector<pair<string, string>> pairs) { if (words1.size() != words2.size()) return false; unordered_map<string, unordered_set<string>> similar_words; for (const auto& pair : pairs) { similar_words[pair.first].insert(pair.second); similar_words[pair.second].insert(pair.first); } for (int i = 0; i < words1.size(); ++i) { if (words1[i] == words2[i]) continue; if (!similar_words[words1[i]].count(words2[i])) return false; } return true; } }; |
Java
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// Author: Huahua // Runtime: 10 ms class Solution { public boolean areSentencesSimilar(String[] words1, String[] words2, String[][] pairs) { if (words1.length != words2.length) return false; Map<String, Set<String>> similar_words = new HashMap<>(); for (String[] pair : pairs) { if (!similar_words.containsKey(pair[0])) similar_words.put(pair[0], new HashSet<>()); if (!similar_words.containsKey(pair[1])) similar_words.put(pair[1], new HashSet<>()); similar_words.get(pair[0]).add(pair[1]); similar_words.get(pair[1]).add(pair[0]); } for (int i = 0; i < words1.length; ++i) { if (words1[i].equals(words2[i])) continue; if (!similar_words.containsKey(words1[i])) return false; if (!similar_words.get(words1[i]).contains(words2[i])) return false; } return true; } } |
Python
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""" Author: Huahua Runtime: 62 ms """ class Solution: def areSentencesSimilar(self, words1, words2, pairs): if len(words1) != len(words2): return False similar_words = {} for w1, w2 in pairs: if not w1 in similar_words: similar_words[w1] = set() if not w2 in similar_words: similar_words[w2] = set() similar_words[w1].add(w2) similar_words[w2].add(w1) for w1, w2 in zip(words1, words2): if w1 == w2: continue if w1 not in similar_words: return False if w2 not in similar_words[w1]: return False return True |
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