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”]
返回words中和pattern匹配的单词list。 和pattern匹配是指相应的字母匹配。多个字母不能同时匹配一个字母。
： 可以用hashmap保存映射关系，但我看到了一种更简单的方法。
如果单词的字母从左到右匹配pattern字母， 比如mee和pattern “abb” m与a匹配，第一个e和第一个b匹配，它们index是相同的； 然后到了后面的e，找到它出现的第一个index, 第一个必须和b出现的index一样， 这个index相等于在hashmap找到了相应的关系。
用index模拟hashmap, 这样就不需要再存一个了hashmap.
public List<String> findAndReplacePattern(String[] words, String pattern) {
List<String> res = new ArrayList<>(); for(String word : words) {
if(check(word, pattern)) res.add(word); } return res; } boolean check(String word, String pattern) {
for(int i = 0; i < word.length(); i ) {
if(wordspan class="token punctuation">.indexOf(word.charAt(i)) != pattern.indexOf(pattern.charAt(i))) return false;
}
return true;
}