leetcode讲解--890. Find and Replace Pattern

890. Find and Replace Pattern

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

题目地址

这个题的难点在于两个字母的一一对应关系，显然map只能做到单向的映射，我的做法是使用两个map。

java 代码

class Solution {
    public List<String> findAndReplacePattern(String[] words, String pattern) {
        List<String> result = new ArrayList<>();
        for(String s:words){
            char[] w = s.toCharArray();
            char[] p = pattern.toCharArray();
            Map<Character, Character> map1 = new HashMap<>();
            Map<Character, Character> map2 = new HashMap<>();
            for(int i=0;i<w.length;i++){
                if(map1.get(w[i])==null && map2.get(p[i])==null){
                    map1.put(w[i], p[i]);
                    map2.put(p[i], w[i]);
                }else if(map1.get(w[i])!=null && map2.get(p[i])!=null && map1.get(w[i])==p[i] && map2.get(p[i])==w[i]){

                }else{
                    break;
                }
                if(i==w.length-1){
                    System.out.println(s);
                    result.add(s);
                }
            }
        }
        return result;
    }
}