描述
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"].
1.The length of words1 and words2 will not exceed 1000.
2.The length of pairs will not exceed 2000.
3.The length of each pairs[i] will be 2.
4.The length of each words[i] and pairs[i][j] will be in the range [1, 20].
第一想法就是要建立哈希表，但存在一对多的情况，我在考虑使用unordered_multimap ，这个思路以后再写。
用自己最熟悉的unordered_map只不过value值需要是一个set类型，因为会出现一对多的情况 。
class Solution {
public:
/**
* @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
*/
bool isSentenceSimilarity(vector<string> &words1, vector<string> &words2, vector<vector<string>> &pairs) {
// write your code here
if(words1.size()!=words2.size()) return false;
unordered_map<string,unordered_set<string>> m_map;
for(auto pair:pairs){
m_map[pair[0]].insert(pair[1]);
}
for(int i=0;i<words1.size();++i){
if(words1[i]==words2[i]) continue;
if(!m_map[words1[i]].count(words2[i])&&!m_map[words2[i]].count(words1[i])) return false;
}
return true;
}
};