LeetCode: 890. Find and Replace Pattern

题目描述

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

解题思路

对每个字符根据其出现顺序进行编号，然后将字符串用编号表示。如果编号表示的字符串和模式串相吻合，则符合模式。

AC 代码

class Solution {
    // 根据字符出现的顺序对字符编号，并将原字符串用编号表示
    vector<int> getPatternVec(string pattern)
    {
        string chars;
        vector<int> ans;
        for(char x : pattern)
        {
            if(chars.find_first_of(x) == chars.npos)
            {
                chars.push_back(x);
            }
        }

        for(char x : pattern)
        {
            ans.push_back(chars.find_first_of(x));
        }

        return ans;
    }
public:
    vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
        vector<string> ans;
        vector<int> patternVec = getPatternVec(pattern);
        for(size_t i = 0; i < words.size(); ++i)
        {
            if(getPatternVec(words[i]) == patternVec)
            {
                ans.push_back(words[i]);
            }
        }

        return ans;
    }
};