890. Find and Replace Pattern**
https://leetcode.com/problems/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
C++ 实现 1
要完全符合 Pattern, 需要 word
和 pattern
一一映射. 这说明如果 f(x) -> p
那么 g(p) -> x
. 因此使用两个哈希表分别模拟 f
和 g
.
class Solution {
private:
bool isMatch(const string &pattern, const string &s) {
if (s.size() != pattern.size()) return false;
unordered_map<char, char> s2p, p2s;
for (int i = 0; i < s.size(); ++ i) {
if (s2p.count(s[i]) && s2p[s[i]] != pattern[i]) return false;
if (p2s.count(pattern[i]) && p2s[pattern[i]] != s[i]) return false;
s2p[s[i]] = pattern[i];
p2s[pattern[i]] = s[i];
}
return true;
}
public:
vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
vector<string> res;
for (auto &w : words) {
if (isMatch(pattern, w))
res.push_back(w);
}
return res;
}
};