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 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].
给定两个句子，以单词数组的形式给出words1和words2，以及一组相似单词对pairs，判断两个句子是否相似。两个句子的单词数量要一样，而且words1与words2中的单词是两两相似。单词对的相似具有互逆性但是没有传递性。
解法：哈希表HashMap。先对pairs里面的相似单词进行统计，记得两个单词都最为key添加一次。然后在同时循环两个单词数组单词1和单词2，如果单词相同或者在哈希表里有单词1并且值是单词2，就是相似，直到循环结束返回True。否则返还False。
Java:
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;
} Java:
public boolean areSentencesSimilar(String[] words1, String[] words2, String[][] pairs) {
if (words1.length != words2.length) return false;
Map<String, Set<String>> map = new HashMap<>();
for (String[] pair : pairs) {
if (!map.containsKey(pair[0])) {
map.put(pair[0], new HashSet<>());
}
if (!map.containsKey(pair[1])) {
map.put(pair[1], new HashSet<>());
}
map.get(pair[0]).add(pair[1]);
map.get(pair[1]).add(pair[0]);
}
for (int i = 0; i < words1.length; i++) {
if (!words1[i].equals(words2[i]) && (!map.containsKey(words1[i]) || !map.get(words1[i]).contains(words2[i]))) {
return false;
}
// if (words1[i].equals(words2[i])) continue;
// if (map.containsKey(words1[i]) && map.get(words1[i]).contains(words2[i])) continue;
// return false;
}
return true;
} Python:
class Solution(object):
def areSentencesSimilar(self, words1, words2, pairs):
"""
:type words1: List[str]
:type words2: List[str]
:type pairs: List[List[str]]
:rtype: bool
"""
if len(words1) != len(words2): return False
similars = collections.defaultdict(set)
for w1, w2 in pairs:
similars[w1].add(w2)
similars[w2].add(w1)
for w1, w2 in zip(words1, words2):
if w1 != w2 and w2 not in similars[w1]:
return False
return True
C++:
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;
}
};
C++:
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>> m;
for (auto pair : pairs) {
m[pair.first].insert(pair.second);
}
for (int i = 0; i < words1.size(); ++i) {
if (words1[i] == words2[i]) continue;
if (!m[words1[i]].count(words2[i]) && !m[words2[i]].count(words1[i])) return false;
}
return true;
}
};
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[LeetCode] 737. Sentence Similarity II 句子相似度 II