[Leetcode] 734. Sentence Similarity 解题报告

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, "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].

  思路：

  我们可以根据pairs中的内容，建立起字符串到字符串之间的映射，不过需要注意到一个字符串可能会有多个字符串和它类似。为了减少存储空间，我在建立映射的过程中，约定将大的字符串映射到小的字符串中。

  当有了字典之后，我们就依次遍历words1和words2中的对应字符串，看看他们的相似关系是否可以在字典中找到。如果找不到就提前返回false，否则最后返回true。

  代码：

  class Solution {
  public:
      bool areSentencesSimilar(vector<string>& words1, vector<string>& words2, vector<pair<string, string>> pairs) {
          if (words1.size() != words2.size()) {
              return false;
          }
          unordered_map<string, set<string>> hash;     // map from big word to small word
          for (auto it = pairs.begin(); it != pairs.end(); ++it) {
              if (it->first < it->second) {
                  hash[it->second].insert(it->first);
              }
              else {
                  hash[it->first].insert(it->second);
              }
          }
          for (int i = 0; i < words1.size(); ++i) {
              if (words1[i] < words2[i]) {
                  if (hash.count(words2[i]) == 0 || hash[words2[i]].find(words1[i]) == hash[words2[i]].end()) {
                      return false;
                  }
              }
              else if (words1[i] > words2[i]) {
                  if (hash.count(words1[i]) == 0 || hash[words1[i]].find(words2[i]) == hash[words1[i]].end()) {
                      return false;
                  }
              }
          }
          return true;
      }
  };