# 737. Sentence Similarity II

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]`.

a -> {b}

b -> {a, c}

c -> {b, d}

d -> {c}

``````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, unordered_set<string>> m;
for (auto pair : pairs) {
m[pair.first].insert(pair.second);
m[pair.second].insert(pair.first);
}
for (int i = 0; i < words1.size(); ++i) {
if (words1[i] == words2[i]) continue;
unordered_set<string> visited;
queue<string> q{{words1[i]}};
bool succ = false;
while (!q.empty()) {
auto t = q.front(); q.pop();
if (m[t].count(words2[i])) {
succ = true; break;
}
visited.insert(t);
for (auto a : m[t]) {
if (!visited.count(a)) q.push(a);
}
}
if (!succ) return false;
}
return true;
}
};
``````

``````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, unordered_set<string>> m;
for (auto pair : pairs) {
m[pair.first].insert(pair.second);
m[pair.second].insert(pair.first);
}
for (int i = 0; i < words1.size(); ++i) {
unordered_set<string> visited;
if (!helper(m, words1[i], words2[i], visited)) return false;
}
return true;
}
bool helper(unordered_map<string, unordered_set<string>>& m, string& cur, string& target, unordered_set<string>& visited) {
if (cur == target) return true;
visited.insert(cur);
for (string word : m[cur]) {
if (!visited.count(word) && helper(m, word, target, visited)) return true;
}
return false;
}
};
``````

``````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> m;
for (auto pair : pairs) {
string x = getRoot(pair.first, m), y = getRoot(pair.second, m);
if (x != y) m[x] = y;
}
for (int i = 0; i < words1.size(); ++i) {
if (getRoot(words1[i], m) != getRoot(words2[i], m)) return false;
}
return true;
}
string getRoot(string word, unordered_map<string, string>& m) {
if (!m.count(word)) m[word] = word;
return word == m[word] ? word : getRoot(m[word], m);
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/737

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Sentence Similarity

https://leetcode.com/problems/sentence-similarity-ii/

https://leetcode.com/problems/sentence-similarity-ii/discuss/109747/Java-Easy-DFS-solution-with-Explanation

https://leetcode.com/problems/sentence-similarity-ii/discuss/109752/JavaC%2B%2B-Clean-Code-with-Explanation

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