- Find and Replace PatternYou 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 <= 501 <= 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; }}