Leetcode NO.249 Group Shifted Strings

本题题目要求如下：

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"]
]

本题我觉得难度并不低，感觉差不多应该是medium的难度。。或者是我的处理有点复杂。。

我的思路如下：

• 如何找到属于同一组的字符串，使用hashmap，设定val为该组的所有元素，key需要计算一下：比如"abc"，key就是('a' - 'b'), ('b' - 'c')，然后再把这两个数变成字母，这样就做出了key，同理"bcd"也会得到同样的key
• 做完hashmap后，要把每个val都sort，然后新建一个二维vector，即可得到最后的答案，代码如下：

class Solution {
public:
    vector> groupStrings(vector& strings) {
        unordered_map> hashmap;
        vector> res;
        
        for (int i = 0; i < strings.size(); ++i) {
            hashmap[calKey(strings[i])].push_back(strings[i]);
        }
        
        for (auto it = hashmap.begin(); it != hashmap.end(); ++it) {
            sort(it->second.begin(), it->second.end());
            vector tmp = it->second;
            res.push_back(tmp);
        }
        
        return res;
    }
private:
    string calKey(string input) {
        string key = "";
        for (int i = 0; i < input.length() - 1; ++i) {
            int gap = (input[i] > input[i+1]) ? (input[i] - input[i+1]) : (input[i] - input[i+1] + 26);
            char num = 'a' + gap;
            key += num;
        }
        return key;
    }
};