290.Word Pattern

tags: Easy,Hash Table,String

290. 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)))

Yen-Chi ChenSun, 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…
MarsgoatSun, 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= =
Yen-Chi ChenMon, Jan 1, 2023

Reference

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