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737. 句子相似性 II 🔒

题目描述

我们可以将一个句子表示为一个单词数组，例如，句子 I am happy with leetcode"可以表示为 arr = ["I","am",happy","with","leetcode"]

给定两个句子 sentence1 和 sentence2 分别表示为一个字符串数组，并给定一个字符串对 similarPairs ，其中 similarPairs[i] = [xi, yi] 表示两个单词 xi 和 yi 是相似的。

如果 sentence1 和 sentence2 相似则返回 true ，如果不相似则返回 false 。

两个句子是相似的，如果:

  • 它们具有 相同的长度 (即相同的词数)
  • sentence1[i] 和 sentence2[i] 是相似的

请注意，一个词总是与它自己相似，也请注意，相似关系是可传递的。例如，如果单词 a 和 b 是相似的，单词 b 和 c 也是相似的，那么 a 和 c 也是 相似 的。

 

示例 1:

输入: sentence1 = ["great","acting","skills"], sentence2 = ["fine","drama","talent"], similarPairs = [["great","good"],["fine","good"],["drama","acting"],["skills","talent"]]
输出: true
解释: 这两个句子长度相同，每个单词都相似。

示例 2:

输入: sentence1 = ["I","love","leetcode"], sentence2 = ["I","love","onepiece"], similarPairs = [["manga","onepiece"],["platform","anime"],["leetcode","platform"],["anime","manga"]]
输出: true
解释: "leetcode" --> "platform" --> "anime" --> "manga" --> "onepiece".
因为“leetcode”和“onepiece”相似，而且前两个单词是相同的，所以这两句话是相似的。

示例 3:

输入: sentence1 = ["I","love","leetcode"], sentence2 = ["I","love","onepiece"], similarPairs = [["manga","hunterXhunter"],["platform","anime"],["leetcode","platform"],["anime","manga"]]
输出: false
解释: “leetcode”和“onepiece”不相似。

 

提示:

  • 1 <= sentence1.length, sentence2.length <= 1000
  • 1 <= sentence1[i].length, sentence2[i].length <= 20
  • sentence1[i] 和 sentence2[i] 只包含大小写英文字母
  • 0 <= similarPairs.length <= 2000
  • similarPairs[i].length == 2
  • 1 <= xi.length, yi.length <= 20
  • xi 和 yi 只含英文字母

解法

方法一

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class Solution:
    def areSentencesSimilarTwo(
        self, sentence1: List[str], sentence2: List[str], similarPairs: List[List[str]]
    ) -> bool:
        if len(sentence1) != len(sentence2):
            return False
        n = len(similarPairs)
        p = list(range(n << 1))

        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        words = {}
        idx = 0
        for a, b in similarPairs:
            if a not in words:
                words[a] = idx
                idx += 1
            if b not in words:
                words[b] = idx
                idx += 1
            p[find(words[a])] = find(words[b])

        for i in range(len(sentence1)):
            if sentence1[i] == sentence2[i]:
                continue
            if (
                sentence1[i] not in words
                or sentence2[i] not in words
                or find(words[sentence1[i]]) != find(words[sentence2[i]])
            ):
                return False
        return True

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class Solution {
    private int[] p;

    public boolean areSentencesSimilarTwo(
        String[] sentence1, String[] sentence2, List<List<String>> similarPairs) {
        if (sentence1.length != sentence2.length) {
            return false;
        }
        int n = similarPairs.size();
        p = new int[n << 1];
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
        }
        Map<String, Integer> words = new HashMap<>();
        int idx = 0;
        for (List<String> e : similarPairs) {
            String a = e.get(0), b = e.get(1);
            if (!words.containsKey(a)) {
                words.put(a, idx++);
            }
            if (!words.containsKey(b)) {
                words.put(b, idx++);
            }
            p[find(words.get(a))] = find(words.get(b));
        }
        for (int i = 0; i < sentence1.length; ++i) {
            if (Objects.equals(sentence1[i], sentence2[i])) {
                continue;
            }
            if (!words.containsKey(sentence1[i]) || !words.containsKey(sentence2[i])
                || find(words.get(sentence1[i])) != find(words.get(sentence2[i]))) {
                return false;
            }
        }
        return true;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

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class Solution {
public:
    vector<int> p;
    bool areSentencesSimilarTwo(vector<string>& sentence1, vector<string>& sentence2, vector<vector<string>>& similarPairs) {
        if (sentence1.size() != sentence2.size())
            return false;
        int n = similarPairs.size();
        p.resize(n << 1);
        for (int i = 0; i < p.size(); ++i)
            p[i] = i;
        unordered_map<string, int> words;
        int idx = 0;
        for (auto e : similarPairs) {
            string a = e[0], b = e[1];
            if (!words.count(a))
                words[a] = idx++;
            if (!words.count(b))
                words[b] = idx++;
            p[find(words[a])] = find(words[b]);
        }
        for (int i = 0; i < sentence1.size(); ++i) {
            if (sentence1[i] == sentence2[i])
                continue;
            if (!words.count(sentence1[i]) || !words.count(sentence2[i]) || find(words[sentence1[i]]) != find(words[sentence2[i]]))
                return false;
        }
        return true;
    }

    int find(int x) {
        if (p[x] != x)
            p[x] = find(p[x]);
        return p[x];
    }
};

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var p []int

func areSentencesSimilarTwo(sentence1 []string, sentence2 []string, similarPairs [][]string) bool {
    if len(sentence1) != len(sentence2) {
        return false
    }
    n := len(similarPairs)
    p = make([]int, (n<<1)+10)
    for i := 0; i < len(p); i++ {
        p[i] = i
    }
    words := make(map[string]int)
    idx := 1
    for _, e := range similarPairs {
        a, b := e[0], e[1]
        if words[a] == 0 {
            words[a] = idx
            idx++
        }
        if words[b] == 0 {
            words[b] = idx
            idx++
        }
        p[find(words[a])] = find(words[b])
    }
    for i := 0; i < len(sentence1); i++ {
        if sentence1[i] == sentence2[i] {
            continue
        }
        if words[sentence1[i]] == 0 || words[sentence2[i]] == 0 || find(words[sentence1[i]]) != find(words[sentence2[i]]) {
            return false
        }
    }
    return true
}

func find(x int) int {
    if p[x] != x {
        p[x] = find(p[x])
    }
    return p[x]
}

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