Find-and-Replace-Pattern

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题目地址

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:

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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

  判断字符串是否满足 pattern 只需要判断每个字母相对于前边的字母是否相等的规律与 pattern 能否一致就可以了。

解题代码

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class Solution {
public:
    vector<string> findAndReplacePattern(vector<string> &words, string pattern) {
        vector<string> res(0);
        bool isPattern;
        for (auto &word : words) {
            isPattern = true;
            for (int i = 1; i < pattern.size(); ++i) {
                for (int j = 0; j < i; ++j) {
                    if ((word[i] == word[j]) ^ (pattern[i] == pattern[j])) {
                        isPattern = false;
                        break;
                    }
                }
                if (!isPattern) break;
            }
            if (isPattern) res.push_back(word);
        }
        return res;
    }
};
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