290.Word Pattern
===

###### tags: `Easy`,`Hash Table`,`String`

[290. Word Pattern](https://leetcode.com/problems/word-pattern/)

### 題目描述

Given a `pattern` and a string `s`, find if `s` follows the same pattern.

Here **follow** means a full match, such that there is a bijection between a letter in `pattern` and a **non-empty** word in `s`.

### 範例

**Example 1:**

Input: pattern = "abba", s = "dog cat cat dog"
Output: true

**Example 2:**

Input: pattern = "abba", s = "dog cat cat fish"
Output: false

**Example 3:**

Input: pattern = "aaaa", s = "dog cat cat dog"
Output: false

**Constraints**:

* 1 <= `pattern.length` <= 300
* `pattern` contains only lower-case English letters.
* 1 <= `s.length` <= 3000
* `s` contains only lowercase English letters and spaces `' '`.
* `s` **does not contain** any leading or trailing spaces.
* All the words in s are separated by a **single space**.

### 解答
#### Python

class Solution:
    def wordPattern(self, pattern: str, s: str) -> bool:
        words = s.split()
        return len(set(pattern)) == len(set(words)) == len(set(zip_longest(pattern, words)))

> [name=Yen-Chi Chen][time=Sun, Jan 1, 2023]

#### Javascript

function wordPattern(pattern, s) {
  const words = s.split(' ');
  const map = new Map();
  if (words.length !== pattern.length) return false;

  for (let i = 0; i < pattern.length; i++) {
    if (map.has(pattern[i])) {
      if (map.get(pattern[i]) !== words[i]) return false;
    } else {
      if ([...map.values()].includes(words[i])) return false;
      map.set(pattern[i], words[i]);
    }
  }

  return true;
}

> 看到樓上的一行超人，羞愧得差點不敢發...
> [name=Marsgoat][time=Sun, Jan 1, 2023]

#### Python

class Solution:
    def wordPattern(self, pattern: str, s: str) -> bool:
        return len(set(pattern)) == len(set(words:=s.split())) == len(set(zip_longest(pattern, words)))

> 這才是一行好嗎= =
> [name=Yen-Chi Chen][time=Mon, Jan 1, 2023]
### Reference

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