共计 1411 个字符，预计需要花费 4 分钟才能阅读完成。
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;
}
}
