[LeetCode249]Group Shifted Strings

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.

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这种group xxx的题， 都想到hashtable key是这些string 的共同特点，vector里就装这些string自己。
本题的key就是这些字母之间的关系，shift无非就是每个字母之间的gap是一样的。
比如”abc”, “bcd”, “xyz”之间都是1。所以我把它们对应“011“这个key。
但是， 注意到”az” and “ba”是一个group的，按照上面的分析，az应该对应26－1 ＝ 25， ba对应 1－2 ＝ －1。如何让－1 ＝＝ 25？加个26啊。。。
所以code如下：

class Solution {
public:
    vector<vector<string>> groupStrings(vector<string>& strings) {
        unordered_map<string, vector<string>> mp;
        for(string s : strings){
            mp[preProcess(s)].push_back(s);
        }
        vector<vector<string>> res;
        for(auto m : mp){
            vector<string> tmp = m.second;
            sort(tmp.begin(), tmp.end());
            res.push_back(tmp);
        }
        return res;
    }
    string preProcess(string s){
        string res;
        for (int i = 1; i<s.size(); ++i) {
            int tmp = s[i] - s[i-1] >= 0 ? s[i] - s[i-1] : s[i] - s[i-1]+26;
            res += 'a' + tmp;
        }
        return res;
    }
};