0890. Find and Replace Pattern - kumaeki/LeetCode GitHub Wiki

0890. Find and Replace Pattern

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"]

Constraints:

  • 1 <= pattern.length <= 20
  • 1 <= words.length <= 50
  • words[i].length == pattern.length
  • pattern and words[i] are lowercase English letters.

解法1

class Solution {
    public List<String> findAndReplacePattern(String[] words, String pattern) {
        List<String> res = new ArrayList<String>();
        for(String w : words)
            if(helper(w, pattern) && helper(pattern, w))
                res.add(w);
        
        return res;
    }
    
    // 判断w1和w2能否匹配
    private boolean helper(String w1, String w2){
        
        // 定义一个map来保存字符的映射
        Map<Character, Character> map = new HashMap<Character, Character>();
        int len = w1.length();
        for(int i = 0; i < len; i++){
            // 得到当前的字符 c1, c2
            char c1 = w1.charAt(i), c2 = w2.charAt(i);
            // 如果map 中c1是第一次出现, 将c1 和 c2 放入map中
            if(!map.containsKey(c1))
                map.put(c1, c2);
            
            // 如果存在, 但是上次c1对应的字符不是c2, 那么w 和 p 不匹配
            else if(map.get(c1) != c2)
                return false;
        }
        return true;
    }
}
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