LeetCode 49. Group Anagrams

unordered_map< string, vector<string> >map;存sort之后的每个string及其一串anagram。只是函数的简单调用。注释部分是另一种语法，思路是一样的。

 

class Solution {
public:
    string Sort(string a){
        int len = a.length();
        int cnt[27] = {0};
        for(int i = 0; i < a.length(); ++i){
            ++cnt[a[i] - 'a'];
        }
        string res = "";
        for(int i = 0; i < 27; ++i){
            int tmp = cnt[i];
            while(tmp){
                res += char(i + 'a');
                --tmp;
            }
        }
        return res;
    }
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map< string, vector<string> >map;
        for(int i = 0; i <strs.size(); ++i){
            string copy = strs[i];
            sort(copy.begin(), copy.end());
            // copy = Sort(copy);
            if(map.find(copy) == map.end()){
                vector<string> anagrams;
                anagrams.push_back(strs[i]);
                map[copy] = anagrams;
            }
            else map[copy].push_back(strs[i]);
        }
        
        vector< vector<string> >res;
        for(auto str : map){
            sort(str.second.begin(), str.second.end());
            vector<string> tmp = str.second;
            
            res.push_back(tmp);
        }
        return res;
        
        // for(auto &p : classfication){
        //     sort(classfication[p.first].begin(), classfication[p.first].end());
        //     ans.push_back(classfication[p.first]);
        // }
    }
};