力扣-290.单词规律

Idea

1. 先建立一个hashmap，记录s串中的每个单词以及对应的下标
2. 再建立一个hashmap，记录pattern串中相同字母以及对应的下标
3. 遍历pattern串时，遇到不同字母存到pat表中，同时将下标对应的s中的单词存入到查重test集中，因为如果输入样例为"abba",“dog dog dog dog”，无法得出正确答案
4. 最后通过比较pattern中有相同字母的位置对应s中出现相同的word即可，否则为false

AC Code

class Solution {
public:bool wordPattern(string pattern, string s) {unordered_map<char,int> pat;unordered_map<int,string> word;int n = s.size();int index = 0;for(int i = 0; i < n; i++) {string tmp = "";while(s[i] != ' ' && i < n){tmp += s[i];i++;}word[index++] = tmp;}if(word.size() != pattern.size()) return false;unordered_set<string> test;for(int i = 0; i < pattern.size(); i++) {if(pat.find(pattern[i]) == pat.end()){pat[pattern[i]] = i;if(test.find(word[i]) != test.end()){return false;}test.insert(word[i]);}else {int con1 = pat[pattern[i]];int con2 = i;if(word[con1] != word[con2]) return false;if(pattern[con1] != pattern[con2]) return false;}}return true;}
};