The problem is very tricky, first I use two rules:
1. If two words are anagrams then they have same length
2. If two words are anagrams then if have same map
Then, use loop to traverse the map and do some handle, but the time complexity is O(n^2), which gets TLE at last.

The solution of leetcode is first sort the word then directly append the word to the map, and use the sorted word as key, so that we do not need to traverse the array of map;

Another solution is build a key base on current word in O(n), then directly search the key in the map, which do not need to sort the key and decrease the time.

In summary, the problem is try to decrease O(n^2) to O(nklog^k), which is use the word as key directly to achieve it.

Error:
1. Use wrong solution, which is O(n^2)
2. When reimplement second one, I used map<string, string>, the right one is map<string, vector>.

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Now I think we can use a map like <vector, vector > to store it, which vector is the appearance of 26 alphabet