890. 查找和替换模式

题意

题目链接
判断二个字符串里面的每个字符是否满足一一映射。

思路

直接暴力匹配

代码

class Solution {
public:
    bool judge(const string &word, const string &pattern) {
        map<char, char> mp1, mp2;
        if (word.size() != pattern.size())
            return false;
        for (int i = 0; i < word.size(); i++)
        {
            if (mp1.count(word[i]) == 0)
                mp1[word[i]] = pattern[i];
            else if (mp1[word[i]] != pattern[i])
                return false;
            if (mp2.count(pattern[i]) == 0)
                mp2[pattern[i]] = word[i];
            else if (mp2[pattern[i]] != word[i])
                return false;
        }
        return true;
    }
    vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
        vector<string> ans;
        for (vector<string>::iterator it = words.begin(); it != words.end(); ++it)
            if (judge(*it, pattern))
                ans.push_back(*it);
        return ans;
    }
};