890. Find and Replace Pattern

Problem:

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

思路：

Solution (C++):

vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
    vector<string> res;
    string p = F(pattern);
    for (auto w : words) {
        if (F(w) == p)
            res.push_back(w);
    }
    return res;
}
string F(string w) {
    unordered_map<char, int> m;
    for (auto c : w) 
        if (m.count(c) == 0) 
            m[c] = m.size();
    for (int i = 0; i < w.size(); ++i)
        w[i] = 'a' + m[w[i]];
    return w;
}

性能：

Runtime: 4 ms  Memory Usage: 7.3 MB

思路：

Solution (C++):

性能：

Runtime: ms  Memory Usage: MB