734. Sentence Similarity 有字典数组的相似句子

［抄题］：

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"].

[暴力解法]：

时间分析：

空间分析：

[优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

［一句话思路］：

先存后取

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

可能有、可能没有value对应时，用 .getOrDefault 方法

［二刷］：

［三刷］：

［四刷］：

［五刷］：

[五分钟肉眼debug的结果]：

［总结］：

在多维数组中，p[0] p [1]对应的还是一个元素，不是一坨元素

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［英文数据结构或算法，为什么不用别的数据结构或算法］：

hashset 即使有多次对应关系，也只存一次

［关键模板化代码］：

［其他解法］：

［Follow Up］：

737. Sentence Similarity II union find真不懂为啥

［LC给出的题目变变变］：

[代码风格] ：

在多维数组中，p[0] p [1]对应的还是一个元素，不是一坨元素，纠正一下概念。

class Solution {
    public boolean areSentencesSimilar(String[] words1, String[] words2, String[][] pairs) {
        //cc
        if (words1.length != words2.length) {
            return false;
        }

        //ini, HahsMap<String, new HashSet<String>>
        Map<String, HashSet<String>> map = new HashMap<>();

        //put into hashmap
        for (String p[] : pairs) {
            if (!map.containsKey(p[0])) {
                map.put(p[0], new HashSet<String>());
            }
            map.get(p[0]).add(p[1]);
        }

        //check if a = b, ab, ba
        for (int i = 0; i < words1.length; i++) {
            if (!words1[i].equals(words2[i]) && !(map.getOrDefault(words1[i], new HashSet<String>())).contains(words2[i]) &&
                !(map.getOrDefault(words2[i], new HashSet<String>())).contains(words1[i])) return false;
        }

        return true;
    }
}