[LeetCode] 890. Find and Replace Pattern 查找和替换模式

2021年09月15日 阅读数:4
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You have a list of words and a pattern, and you want to know which words in words matches the pattern.html

A word matches the pattern if there exists a permutation of letters pso that after replacing every letter x in the pattern with p(x), we get the desired word.java

(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.)python

Return a list of the words in words that match the given pattern. 数组

You may return the answer in any order.函数

Example 1:post

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. 


  • 1 <= words.length <= 50
  • 1 <= pattern.length = words[i].length <= 20

给一个单词列表和一个模式字符串,从单词数组中找到和字符串模式匹配的单词。与题目205. Isomorphic Strings 相似spa



public List<String> findAndReplacePattern(String[] words, String pattern) {
        int[] p = F(pattern);
        List<String> res = new ArrayList<String>();
        for (String w : words)
            if (Arrays.equals(F(w), p)) res.add(w);
        return res;

    public int[] F(String w) {
        HashMap<Character, Integer> m = new HashMap<>();
        int n = w.length();
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            m.putIfAbsent(w.charAt(i), m.size());
            res[i] = m.get(w.charAt(i));
        return res;


# Time:  O(n * l)
# Space: O(1)
import itertools

class Solution(object):
    def findAndReplacePattern(self, words, pattern):
        :type words: List[str]
        :type pattern: str
        :rtype: List[str]
        def match(word):
            lookup = {}
            for x, y in itertools.izip(pattern, word):
                if lookup.setdefault(x, y) != y:
                    return False
            return len(set(lookup.values())) == len(lookup.values())

        return filter(match, words)


def findAndReplacePattern(self, words, p):
        def F(w):
            m = {}
            for c in w: m[c] = m.get(c, len(m))
            return "".join(chr(m[c] + 97) for c in w)
        return [w for w in words if F(w) == F(p)] 

Python: Similar to isomorphic string, check the length of the sets and the length of the set when the characters are zipped together.

b = pattern
def is_iso(a):
    return len(a) == len(b) and len(set(a)) == len(set(b)) == len(set(zip(a, b)))
return filter(is_iso, words) 


vector<string> findAndReplacePattern(vector<string> words, string p) {
        vector<string> res;
        for (string w : words) if (F(w) == F(p)) res.push_back(w);
        return res;

    string F(string w) {
        unordered_map<char, int> m;
        for (char c : w) if (!m.count(c)) m[c] = m.size();
        for (int i = 0; i < w.length(); ++i) w[i] = 'a' + m[w[i]];
        return w;


// Time:  O(n * l)
// Space: O(1)
class Solution {
    vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
        vector<string> result;
        for (const auto& word: words) {
            if (match(word, pattern)) {
        return result;

    bool match(const string& word, const string& pattern) {
        unordered_map<char, char> lookup;
        unordered_set<char> char_set;
        for (int i = 0; i < word.length(); ++i) {
            const auto& c = word[i], &p = pattern[i];
            if (!lookup.count(c)) {
                if (char_set.count(p)) {
                    return false;
                lookup[c] = p;
            if (lookup[c] != p) {
                return false;
        return true;



[LeetCode] 205. Isomorphic Strings 同构字符串


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