890. Find and Replace Pattern

BigFatSheep

于 2018-11-22 16:55:42 发布

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分类专栏： Leetcode Medium

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本文介绍了一种高效的模式匹配算法，用于在一组单词中寻找与特定模式相匹配的单词。通过将模式转换为数字表示，可以快速判断单词是否符合相同的模式结构。

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Description

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

Problem URL

Solution

给一组words和一个pattern，找到words中所有的符合这个pattern的word。pattern类似于abb这个样子，相同的就是mee，cdd。

Convert pattern to a fixed using number to denote the character in this positon. Then find wether word and pattern have same pattern.

Code

class Solution {
    public List<String> findAndReplacePattern(String[] words, String pattern) {
        int[] p = find(pattern);
        List<String> res = new ArrayList<>();
        for (String word : words){
            if (Arrays.equals(find(word), p)){
                res.add(word);
            }
        }
        return res;
    }
    
    private int[] find(String word){
        HashMap<Character, Integer> map = new HashMap<>();
        int n = word.length();
        int[] res = new int[n];
        for (int i = 0; i < n; i++){
            if (!map.containsKey(word.charAt(i))){
                map.put(word.charAt(i), map.size());
            }
            res[i] = map.get(word.charAt(i));
        }
        return res;
    }
}