249. Group Shifted Strings --------M

Given a string, we can "shift" each of its letter to its successive letter, for example: "abc" -> "bcd". We can keep "shifting" which forms the sequence:

"abc" -> "bcd" -> ... -> "xyz"

Given a list of strings which contains only lowercase alphabets, group all strings that belong to the same shifting sequence.

For example, given: ["abc", "bcd", "acef", "xyz", "az", "ba", "a", "z"],
Return:

[
  ["abc","bcd","xyz"],
  ["az","ba"],
  ["acef"],
  ["a","z"]
]

Note: For the return value, each inner list's elements must follow the lexicographic order.

A:

class Solution {
public:
    vector<vector<string>> groupStrings(vector<string>& strings) {
        unordered_map<string, vector<string>> M;
        for(auto &s:strings){ // represent each string with their char difference
            string tmp="";
            for(int i =1;i<s.length();i++){
                tmp+= to_string( (26 + s[i]-s[i-1]) %26);
            }
            M[tmp].push_back(s);
        }
        vector<vector<string>> res;
        auto iter = M.begin();
        while(iter!= M.end()){
            res.push_back(iter->second);
            iter++;
        }
        return res;
    }
};

1:  一个整数 -1 % 26,  的结果为 -1  ， 因此计算新的char的时候， 得加上26

一开始犯了糊涂，  az,  za 的difference 的结果是不一样的。