[LeetCode] Sentence Similarity

Posted 2020-10-16 immjc

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].

判断句子相似性。
给定两个字符串数组和一个由键值对组成的数组。
根据给定的规则判断字符串数组对应的字符串是否相等。即键值对中的键值是否对应于字符串数组中的对应字符串。
首先先判断边界条件，三种特殊情况。
然后外层for循环遍历字符串数组，设置标志位，内层for循环判断键值对中的字符是否与字符串数组中对应的字符串是否相等。其中还需要判断两个字符串相等的情况
每循环一次内层for前将标志位置为false。如果内层for遍历结束标志位仍为false。则认为不相似返回false即可。
最后结束遍历 后返回true。

class Solution {
public:
    bool areSentencesSimilar(vector<string>& words1, vector<string>& words2, vector<pair<string, string>> pairs) {
        if (words1.empty() || words2.empty())
            return true;
        if (pairs.empty())
            return true;
        if (words1.size() != words2.size())
            return false;
        
        bool isFind;
        for (int i = 0; i < words1.size(); i++) {
            isFind = false;
            for (int j = 0; j < pairs.size(); j++) {
                if (words1[i] == pairs[j].first && words2[i] == pairs[j].second)
                    isFind = true;
                if (words1[i] == pairs[j].second && words2[i] == pairs[j].first)
                    isFind = true;
                if (words1[i] == words2[i])
                    isFind = true;
            }
            if (!isFind)
                return false;
        }
        return true;
    }
};
// 6 ms