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
andwords2
will not exceed1000
. - The length of
pairs
will 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]
.
思路:hashMap<String, HashSet<String>>因为有一个单词跟很多单词类似,那么就是一个临接set,注意要check双向关系,不能只check一边;T(N) S(N)
class Solution {
public boolean areSentencesSimilar(String[] words1, String[] words2, List<List<String>> pairs) {
if(words1 == null && words2 == null) {
return true;
}
if(words1 == null || words2 == null || words1.length != words2.length) {
return false;
}
HashMap<String, HashSet<String>> hashmap = new HashMap<>();
for(List<String> pair: pairs) {
String from = pair.get(0);
String to = pair.get(1);
hashmap.putIfAbsent(from, new HashSet<String>());
hashmap.putIfAbsent(to, new HashSet<String>());
hashmap.get(from).add(to);
hashmap.get(to).add(from);
}
for(int i = 0; i < words1.length; i++) {
String word1 = words1[i];
String word2 = words2[i];
if(word1.equals(word2)) {
continue;
}
if(hashmap.containsKey(word1)) {
if(!hashmap.get(word1).contains(word2)) {
return false;
}
}
if(hashmap.containsKey(word2)) {
if(!hashmap.get(word2).contains(word1)) {
return false;
}
}
}
return true;
}
}