LC 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.

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

Runtime: 4 ms, faster than 62.45% of C++ online submissions for Find and Replace Pattern.

注意一一对应要用两个字典。

class Solution {
public:
  vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
    vector<string> ret;
    for(auto word : words){
      if(word.size() != pattern.size()) continue;
      unordered_map<char,char> mp_w2p;
      unordered_map<char,char> mp_p2w;
      bool shouldput = true;
      for(int i=0; i<word.size(); i++){
        if(!mp_w2p.count(word[i]) && !mp_p2w.count(pattern[i])){
          mp_w2p[word[i]] = pattern[i];
          mp_p2w[pattern[i]] = word[i];
        } else if(!mp_w2p.count(word[i]) || !mp_p2w.count(pattern[i])) {
          shouldput = false;
          break;
        }else if(mp_w2p[word[i]] != pattern[i] || mp_p2w[pattern[i]] != word[i]){
          shouldput = false;
          break;
        }
      }
      if(shouldput) ret.push_back(word);
    }
    return ret;
  }
};

原文地址：https://www.cnblogs.com/ethanhong/p/10188208.html