737. Sentence Similarity II
Problem:
The length of
The length of
The length of each
The length of each
Analysis:
Before reading sentence similarity II, it was a little bit difficult to understand the problem. If 2 words are similar, qualify the following conditions:
1. Length equal
2. words1[i] == words2[i] || words1[i] is similar to words2[i]
Typical Union find problem. Use hashmap to convert string to integer.
Solution:
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:
words1
and words2
will not exceed 1000
.pairs
will not exceed 2000
.pairs[i]
will be 2
.words[i]
and pairs[i][j]
will be in the range [1, 20]
.Before reading sentence similarity II, it was a little bit difficult to understand the problem. If 2 words are similar, qualify the following conditions:
1. Length equal
2. words1[i] == words2[i] || words1[i] is similar to words2[i]
Typical Union find problem. Use hashmap to convert string to integer.
Solution:
class Solution { class UnionFind { int[] father; public UnionFind(int n) { father = new int[n]; for (int i = 0; i < n; i++) { father[i] = i; } } public int find(int n) { if (father[n] != n) father[n] = find(father[n]); return father[n]; } public void union(int a, int b) { father[find(a)] = find(b); } public boolean query(int a, int b){ return find(a) == find(b); } } public boolean areSentencesSimilarTwo(String[] words1, String[] words2, String[][] pairs) { if (words1 == null || words2 == null || words1.length != words2.length || pairs == null) return false; if (pairs.length == 0) return true; UnionFind uf = new UnionFind(pairs.length * 2); Map<String, Integer> index = new HashMap<>(); int count = 0; for (String[] pair: pairs) { for (String p: pair) { if (!index.containsKey(p)) { index.put(p, count++); } } uf.union(index.get(pair[0]), index.get(pair[1])); } for (int i = 0; i < words1.length; i++) { String w1 = words1[i]; String w2 = words2[i]; if (w1.equals(w2)) continue; if (!index.containsKey(w1) || !index.containsKey(w2) || uf.find(index.get(w1)) != uf.find(index.get(w2))) { return false; } } return true; } }
评论
发表评论