[Leetcode] 737. Sentence Similarity II 解题报告

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魔豆Magicbean

最新推荐文章于 2024-05-10 23:40:00 发布

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分类专栏： IT公司面试习题文章标签： Leetcode 解题报告 union-find

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这篇博客介绍了如何解决LeetCode上的737题——Sentence Similarity II，通过分析相似词对，探讨两个句子在考虑对称性和传递性的相似性条件下的判断方法。

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题目：

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, words1 = ["great", "acting", "skills"] and words2 = ["fine", "drama", "talent"] are similar, if the similar word pairs are pairs = [["great", "good"], ["fine", "good"], ["acting","drama"], ["skills","talent"]].

Note that the similarity relation is transitive. For example, if "great" and "good" are similar, and "fine" and "good" are similar, then "great" and "fine" are similar.

Similarity is also 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].

思路：

由于similar关系具有传递性，所以我们无法再用简单的哈希表来表示similar关系了，但是我们仅仅需要将简单的哈希表变为union-find即可。具体做法是：首先置所有的string的parents为其本身，然后每当遇到一对pair，就将其中的两个string所在的组进行合并（实现中只需要将其中一个string的parent置为另一个string即可）。这样最后在遍历words1和words2中的每个单词的时候，只需要检查对应单词最终的parent是否一样就可以了。

代码：

class Solution {
public:
    bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>> pairs) {
        if (words1.size() != words2.size()) {
            return false;
        }
        unordered_map<string, string> parents;
        for (auto &p : pairs) {
            parents[p.first] = p.first;
            parents[p.second] = p.second;
        }
        for (auto &p : pairs) {
            string s1 = p.first, s2 = p.second;
            while (parents[s1] != s1) {
                s1 = parents[s1];
            }
            while (parents[s2] != s2) {
                s2 = parents[s2];
            }
            parents[s1] = s2;
        }
        for (int i = 0; i < words1.size(); ++i) {
            if (words1[i] == words2[i]) {
                continue;
            }
            string s1 = words1[i], s2 = words2[i];
            while (parents[s1] != s1) {
                s1 = parents[s1];
            }
            while (parents[s2] != s2) {
                s2 = parents[s2];
            }
            if (s1 != s2) {
                return false;
            }
        }
        return true;
    }
};