LeetCode49

两个方法：

1.给单词排序

按照a-z，比如eat，排完序过后就是aet，那么排完序后对应相同字符串的字符串归类到同一个集合里面
{
aet:[“ate”,“eat”,“tea”]
abt:[“bat”]
ant:[“nat”,“tan”]
}

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map<string, vector<string>> mp;
        for (string& str: strs) {
            string key = str;
            sort(key.begin(), key.end());
            mp[key].emplace_back(str);
        }
        vector<vector<string>> ans;
        for (auto it = mp.begin(); it != mp.end(); ++it) {
            ans.emplace_back(it->second);
        }
        return ans;
    }
};

2.哈希法

定义一个长26的数组shuzu，表示a-z是否出现，例如shuzu[0]=1，则表示a出现。
那么eat对应的数组是[1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,…,1,…]
最后对应数组相同的归类为一组。

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        // 自定义对 array<int, 26> 类型的哈希函数
        auto arrayHash = [fn = hash<int>{}] (const array<int, 26>& arr) -> size_t {
            return accumulate(arr.begin(), arr.end(), 0u, [&](size_t acc, int num) {
                return (acc << 1) ^ fn(num);
            });
        };

        unordered_map<array<int, 26>, vector<string>, decltype(arrayHash)> mp(0, arrayHash);
        for (string& str: strs) {
            array<int, 26> counts{};
            int length = str.length();
            for (int i = 0; i < length; ++i) {
                counts[str[i] - 'a'] ++;
            }
            mp[counts].emplace_back(str);
        }
        vector<vector<string>> ans;
        for (auto it = mp.begin(); it != mp.end(); ++it) {
            ans.emplace_back(it->second);
        }
        return ans;
    }
};