# 49. Group Anagrams

## 1.問題

* **給予一個string 的array, 將相同anagrams的字串分組**

![](/files/-LNtrBkCJ-1z9nuyQWbB)

## 2.想法 <a href="#id-2-xiang-fa" id="id-2-xiang-fa"></a>

* 提問
* function header, parameter
* test input
* 說明想法
  * 可以把這題視為分組問題, 因此分組的關鍵在於將相同key的放在一起, 如果對容器的使用足夠熟悉, 會想到用map搭配vector存放同一群組的string
  * key可以是經過sort的string, 或是另一個map
* 測試計算複雜度

## **3.程式碼** <a href="#id-3-cheng-shi" id="id-3-cheng-shi"></a>

```
class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        vector<vector<string>> res;
        
        if (strs.empty()) {
            return res;
        }    
        
        map<string, vector<string>> m1;
        
        for (int i = 0; i < strs.size(); i++) {
            string key = strs[i];
            sort(key.begin(), key.end());
            m1[key].push_back(strs[i]);
        }
        
        for (map<string, vector<string>>::iterator it= m1.begin(); it != m1.end(); it++) {
            res.push_back(it->second);
        }
        return res;
    }
};
```


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