[LeetCode/LintCode] Sentence Similarity

分手后的思念是犯贱 2021-09-18 14:18 200阅读 0赞

### 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 = \["doubleplus","good"\].

### Solution ###

public class Solution {
        /**
         * @param words1: a list of string
         * @param words2: a list of string
         * @param pairs: a list of string pairs
         * @return: return a boolean, denote whether two sentences are similar or not
         */
        public boolean isSentenceSimilarity(String[] words1, String[] words2, List<List<String>> pairs) {
            // write your code here
            if (words1.length != words2.length) return false;
            Map<String, Set<String>> map = new HashMap<>();
            for (List<String> pair: pairs) {
                if (!map.containsKey(pair.get(0))) {
                    map.put(pair.get(0), new HashSet<String>());
                }
                if (!map.containsKey(pair.get(1))) {
                    map.put(pair.get(1), new HashSet<String>());
                }
                map.get(pair.get(0)).add(pair.get(1));
                map.get(pair.get(1)).add(pair.get(0));
            }
            for (int i = 0; i < words1.length; i++) {
                // if (words1[i] == words2[i]) continue;
                if (!map.containsKey(words1[i])) return false;
                if (!map.get(words1[i]).contains(words2[i])) return false;
            }
            return true;
        }
    }