4-5

LeetCode454

class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        int ans=0;
        unordered_map<int,int> um; //设置查找表
        for(int i=0;i<C.size();i++)
        {
            for(int j=0;j<D.size();j++)
            {
                um[C[i]+D[j]]++; //将C中元素和D中元素的和放到表中，表中记录次数
            }
        }
        for(int i=0;i<A.size();i++)
        {
            for(int j=0;j<B.size();j++)
            {
                int temp=-A[i]-B[j];
                if(um.find(temp)!=um.end())  //如果找到temp，说明找到组合相加等于0
                {
                    ans+=um[temp];  //更新组合的个数
                }

            }
        }
        return ans;

    }
};

LeetCode49

方法1：排序

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;
    }
};

方法2：质数相乘

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        vector<vector<string>> res;
        unordered_map <double,vector<string> > m;
        double a[26]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101};
        for(string& s : strs)
        {
            double t = 1;
            for(char c : s)
                t *= a[c - 'a'];
            m[t].push_back(s);          //t为单词对应的质数乘积，m[t]则为该单词的异位词构成的vector
        }
        for(auto& n : m)                //n为键和值组成的pair
            res.push_back(n.second);
        return res;
    }
};

LeetCode447

LeetCode149