标签：word.size perm find with back div another exist eve
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 <= 501 <= 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;
}
};
LC 890. Find and Replace Pattern
标签：word.size perm find with back div another exist eve
原文地址：https://www.cnblogs.com/ethanhong/p/10188208.html
