[LeetCode] 737. Sentence Similarity II 句子相似度 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].

734. Sentence Similarity 的拓展。不同点，这题的相似单词可以传递。

解法1：DFS。本质是无向连通图的问题，把pairs数组中的每一对相似的字符串视为无向图上的两个结点，对每个结点要记录所有和其相连的结点，比如(a, b), (b, c)和(c, d)的映射关系：a -> {b}， b -> {a, c}， c -> {b, d}， d -> {c}。要验证a和d是否相似，从a只能找到b，b可以找到a、c， a访问过，将访问过的结点加入一个集合visited，此时只能访问c，c里面有b、d，找到d，说明a和d相似。

解法2：BFS。与 DFS 类似，只是写成迭代形式。

解法3：Union Find

Java:

class Solution {
    public boolean areSentencesSimilarTwo(String[] words1, String[] words2, String[][] pairs) {
        if (words1.length != words2.length) {
            return false;
        }
        // Build the graph of pairs
        HashMap<String, Set<String>> pairMap = new HashMap<>();
        for (String[] pair : pairs) {
            // Create keys(words in [][]pairs without duplication) and empty set 
            if (!pairMap.containsKey(pair[0])) {
                pairMap.put(pair[0], new HashSet<String>());
            }
            if (!pairMap.containsKey(pair[1])) {
                pairMap.put(pair[1], new HashSet<String>());
            }
            // Add the corresponding pairs to each other
            pairMap.get(pair[0]).add(pair[1]);
            pairMap.get(pair[1]).add(pair[0]);
        }
      
        // Iterate throught each word in both input strings and do DFS search
        for (int i = 0; i < words1.length; i++) {
            // If same, then we don't need to do DFS search
            if (words1[i].equals(words2[i])) continue;
            // If they are not the same and no such strings in the pairs
            if (!pairMap.containsKey(words1[i]) || !pairMap.containsKey(words2[i])) return false;
            // Do DFS search, initialize the set to prevent revisiting. 
            if (!dfs(words1[i], words2[i], pairMap, new HashSet<>())) return false;
        }
        return true;
    }
         
    public boolean dfs(String source, String target, HashMap<String, Set<String>> pairMap, HashSet<String> visited) {
        if (pairMap.get(source).contains(target)) {
            return true;
        }
        // Mark as visited 
        visited.add(source);
        for (String next : pairMap.get(source)) {
            // DFS other connected node, except the mirrowed string 
            if (!visited.contains(next) && next.equals(target) ||
                !visited.contains(next) && dfs(next, target, pairMap, visited)) {
                return true;    
            }
        }
        // We've done dfs still can't find the target 
        return false;
    }
}　　

Java: Union Find

class Solution {
    public boolean areSentencesSimilarTwo(String[] words1, String[] words2, String[][] pairs) {
        if(words1.length!=words2.length) return false;
        Map<String, String> map = new HashMap<>();
        for(String[] pair : pairs){
            String parent0 = find(pair[0], map);
            String parent1 = find(pair[1], map);
            if(!parent0.equals(parent1)) map.put(parent0, parent1);
        }
        int n = words1.length;
        for(int i=0; i<n; i++){
            if (!words1[i].equals(words2[i]) && !find(words1[i], map).equals(find(words2[i], map))) return false;
        }
        return true;
    }
     
    private String find(String word, Map<String, String> map){
        if(!map.containsKey(word)) return word;
        String str = word;
        while(map.containsKey(str)){
            str = map.get(str);
        }
        map.put(word, str);
        return str;
    }
}

Python: DFS

class Solution(object):
    def areSentencesSimilarTwo(self, words1, words2, pairs):
        """
        :type words1: List[str]
        :type words2: List[str]
        :type pairs: List[List[str]]
        :rtype: bool
        """
        if len(words1) != len(words2): return False
        similars = collections.defaultdict(set)
        for w1, w2 in pairs:
            similars[w1].add(w2)
            similars[w2].add(w1)
 
        def dfs(words1, words2, visits):
            for similar in similars[words2]:
                if words1 == similar:
                    return True
                elif similar not in visits:
                    visits.add(similar)
                    if dfs(words1, similar, visits):
                        return True
            return False
 
        for w1, w2 in zip(words1, words2):
            if w1 != w2 and not dfs(w1, w2, set([w2])):
                return False
        return True　　

Python：BFS

class Solution(object):
    def areSentencesSimilarTwo(self, words1, words2, pairs):
        if len(words1) != len(words2): return False
        graph = collections.defaultdict(list)
        for w1, w2 in pairs:
            graph[w1].append(w2)
            graph[w2].append(w1)
 
        for w1, w2 in zip(words1, words2):
            stack, seen = [w1], {w1}
            while stack:
                word = stack.pop()
                if word == w2: break
                for nei in graph[word]:
                    if nei not in seen:
                        seen.add(nei)
                        stack.append(nei)
            else:
                return False
        return True

Python: DSU(Disjoint Set Union)　　

class DSU:
    def __init__(self, N):
        self.par = range(N)
    def find(self, x):
        if self.par[x] != x:
            self.par[x] = self.find(self.par[x])
        return self.par[x]
    def union(self, x, y):
        self.par[self.find(x)] = self.find(y)
 
class Solution(object):
    def areSentencesSimilarTwo(self, words1, words2, pairs):
        if len(words1) != len(words2): return False
 
        index = {}
        count = itertools.count()
        dsu = DSU(2 * len(pairs))
        for pair in pairs:
            for p in pair:
                if p not in index:
                    index[p] = next(count)
            dsu.union(index[pair[0]], index[pair[1]])
 
        return all(w1 == w2 or
                   w1 in index and w2 in index and
                   dsu.find(index[w1]) == dsu.find(index[w2])
                   for w1, w2 in zip(words1, words2))　　

C++: DFS

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;
    }
};

C++: BFS

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;
    }
};

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, 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);
    }
};