[LeetCode] 890. Find and Replace Pattern

Posted 2021-05-23 CNoodle

Given a list of strings words and a string pattern, return a list of words[i] that match pattern. You may return the answer in any order.

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.

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.

Example 2:

Input: words = ["a","b","c"], pattern = "a"
Output: ["a","b","c"]

Constraints:

• 1 <= pattern.length <= 20
• 1 <= words.length <= 50
• words[i].length == pattern.length
• pattern and words[i] are lowercase English letters.

查找和替换模式。

你有一个单词列表 words 和一个模式  pattern，你想知道 words 中的哪些单词与模式匹配。

如果存在字母的排列 p ，使得将模式中的每个字母 x 替换为 p(x) 之后，我们就得到了所需的单词，那么单词与模式是匹配的。

（回想一下，字母的排列是从字母到字母的双射：每个字母映射到另一个字母，没有两个字母映射到同一个字母。）

返回 words 中与给定模式匹配的单词列表。

你可以按任何顺序返回答案。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-and-replace-pattern
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

这道题类似205题，找同构词，思路一模一样，只不过205题问的是两个词是否互为同构词，这道题问的是 input 数组里面有哪些词跟 pattern 是同构词。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public List<String> findAndReplacePattern(String[] words, String pattern) {
 3         List<String> res = new ArrayList<>();
 4         for (String word : words) {
 5             if (helper(word, pattern)) {
 6                 res.add(word);
 7             }
 8         }
 9         return res;
10     }
11 
12     private boolean helper(String word, String pattern) {
13         HashMap<Character, Character> map = new HashMap<>();
14         for (int i = 0; i < word.length(); i++) {
15             char w = word.charAt(i);
16             char p = pattern.charAt(i);
17             if (!map.containsKey(w)) {
18                 map.put(w, p);
19             }
20             if (map.get(w) != p) {
21                 return false;
22             }
23         }
24 
25         boolean[] seen = new boolean[26];
26         // 因为hashmap存的时候，key是unique的，所以value理应也是unique的
27         // 所以如果出现重复的value就说明是错的
28         for (char p : map.values()) {
29             if (seen[p - \'a\']) {
30                 return false;
31             }
32             seen[p - \'a\'] = true;
33         }
34         return true;
35     }
36 }

 

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890. Find and Replace Pattern

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