1352. Product of the Last K Numbers

 

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].

 

这道题给了我们两个句子，问这两个句子是否是相似的。判定的条件是两个句子的单词数要相同，而且每两个对应的单词要是相似度，这里会给一些相似的单词对，这里说明了单词对的相似具有互逆性但是没有传递性。看到这里博主似乎已经看到了Follow up了，加上传递性就是一个很好的拓展。那么这里没有传递性，就使得问题变得很容易了，我们只要建立一个单词和其所有相似单词的集合的映射就可以了，比如说如果great和fine类似，且great和good类似，那么就有下面这个映射：

great -> {fine, good}

所以我们在逐个检验两个句子中对应的单词时就可以直接去映射中找，注意有可能遇到的单词对时反过来的，比如fine和great，所以我们两个单词都要带到映射中去查找，只要有一个能查找到，就说明是相似的，反之，如果两个都没查找到，说明不相似，直接返回false，参见代码如下：

 

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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