You have a list of
words
and a pattern
, and you want to know which words in words
matches the pattern.
A word matches the pattern if there exists a permutation of letters
p
so that after replacing every letter x
in the pattern with p(x)
, we get the desired word.
(Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.)
Return a list of the words in
words
that match the given pattern.
You may return the answer in any order.
Example 1:
Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb" Output: ["mee","aqq"] Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}. "ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, since a and b map to the same letter.
Note:
1 <= words.length <= 50
1 <= pattern.length = words[i].length <= 20
class Solution { public List<String> findAndReplacePattern(String[] words, String pattern) { List<String> ans = new ArrayList<>(); if (words == null || words.length == 0 || pattern == null || pattern.length() == 0) { return ans; } for (String word : words) { if (match(word, pattern)) { ans.add(word); } } return ans; } private boolean match(String s, String p) { if (s.length() != p.length()) { return false; } Map<Character, Character> map = new HashMap<>(); Map<Character, Character> rMap = new HashMap<>(); for (int i = 0; i < s.length(); i++) { char cs = s.charAt(i); char cp = p.charAt(i); if (!map.containsKey(cs)) { if (rMap.containsKey(cp)) { return false; } map.put(cs, cp); rMap.put(cp, cs); } else { if (map.get(cs) != cp || rMap.get(cp) != cs) { return false; } } } return true; } }Code (Java):
No comments:
Post a Comment