[Leetcode / letter] ectopic packet hash table (hash Map Packet)

Problem Description:

Given a string array, ectopic word letters together. Ectopic word letters refer to the same letters, but arranged in different strings.

Example:

输入: ["eat", "tea", "tan", "ate", "nat", "bat"],
输出:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]

Description:

• All inputs are lowercase letters.
• The order of the answers that were not considered.

The basic idea:

String in ascending order as a key. To be grouped as a vector type value.

AC Code:

class Solution {
public:
    string ChangeOrder(string str) {
    // 返回字符升序排序的字符串
      string res;
      map<char, int> hashmap;
      for (char ch : str) {
        ++hashmap[ch];
      }
      // 按照自然顺序遍历
      for (char ch = 'a'; ch <= 'z'; ++ch) {
        for (int i = 0; i < hashmap[ch]; ++i) {
          res += ch;
        } 
      }
      return res;
    }
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
      map<string, vector<string>> hashmap;
      for (string str : strs) {
        hashmap[ChangeOrder(str)].push_back(str);
      }
      vector<vector<string>> res;
      for (auto it = hashmap.begin(); it != hashmap.end(); ++it) {
        res.push_back(it->second);
      }
      return res;
    }
};

Daniel's ideas:

The above code we spent quite a long time to pass a string copy.

Rather all results stored in the hashmap, then copy the results one by one to the vector.

As a direct operation vector. hashmap is stored in our vector related information.

(Here is the number of rows stored in the vector)

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map<string,int>index;
        vector<vector<string>> result;
        for(auto i:strs)
        {
            //sort(i.begin();i.end());
            string tmp = i;
            sort(tmp.begin(),tmp.end());
            
            if(index.find(tmp)==index.end())
            {
                
                index.insert({tmp,(int)result.size()});
                result.push_back(vector<string>{i});
            }
            else
            {
                result[index[tmp]].push_back(i);
            }

        }
        return result;
    }
};

Other experience:

1.  You may wish to directly operate on the result set returned , so when you create a copy of a waste of space .
2. I sort each character string using a counting sorting method, in fact, we use sort will be more efficient.
3. We can use hashmap to group, in which the key is to mark our group. value is a collection of group members .