Leetcode学习笔记：#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.

实现：

public List<String> findAndReplacePattern(String[] words, String pattern)){
	int[] p = helper(pattern);
	List<String> res = new ArrayList<String>();
	for(String w : words)
		if(Arrays.equals(helper(w), p))
			res.add(w);
	return res;
}

public int[] helper(String w){
	HashMap<Character, Integer> m = new HashMap<>();
	int n = w.length();
	int[] res = new int[n];
	for(int i = 0; i < n; i++){
		m.putIfAbsent(w.charAt(i), m.size());
		res[i] = m.get(w.charAt(i));
	}
	return res;
}
           

思路：

用一个helper方法帮助将string转化为容易判断的数字：运用hashmap中的特性：不能有重复的key，则只要把key的value设成不一样的就很好判断了。例如：abb传入helper 生成数组res[0,1,1]，同样的，poo传入helper也生成数组[0,1,1]。