2023-02-20 leetcode-AccountsMerge

摘要:

记录对leetcode-AccountsMerge的反思

要求:

Given a list accounts, each element accounts[i] is a list of strings, where the first element 
accounts[i][0] is a name, and the rest of the elements are emails representing emails of the 
account.
 * 
Now, we would like to merge these accounts.  Two accounts definitely belong to the same person if 
there is some email that is common to both accounts.  Note that even if two accounts have the same 
name, they may belong to different people as people could have the same name.  A person can have 
any number of accounts initially, but all of their accounts definitely have the same name.
 * 
After merging the accounts, return the accounts in the following format: the first element of each 
account is the name, and the rest of the elements are emails in sorted order.  The accounts 
themselves can be returned in any order.
 * 
Example 1:
 * 
Input: 
accounts = [["John", "[e-mail]", "[e-mail]"], ["John", "[e-mail]"], 
["John", "[e-mail]", "[e-mail]"], ["Mary", "[e-mail]"]]
Output: [["John", '[e-mail]', '[e-mail]', '[e-mail]'],  ["John", 
"[e-mail]"], ["Mary", "[e-mail]"]]
 * 
Explanation: 
The first and third John's are the same person as they have the common email "[e-mail]".
The second John and Mary are different people as none of their email addresses are used by other 
accounts.
We could return these lists in any order, for example the answer [['Mary', '[e-mail]'], 
['John', '[e-mail]'], 
['John', '[e-mail]', '[e-mail]', '[e-mail]']] would still be accepted.
 * 
Note:
The length of accounts will be in the range [1, 1000].
The length of accounts[i] will be in the range [1, 10].
The length of accounts[i][j] will be in the range [1, 30].

数据结构相关:

1. std::vector
2. std::map
3. std::unordered_map

题解:

题解一:

// Declare variables for accounts, mapping and result
vector<vector<string>> accounts;
unordered_map<string, string> mapping;
vector<vector<string>> result;// Add accounts to 'accounts'
// ...// Traverse accounts
for (auto &a : accounts) {// Take the first email address and consider it as the root string root = a[1];for (int i = 2; i < a.size(); i++) {// If current email is already present in map if (mapping.find(a[i]) != mapping.end())root = mapping[a[i]]; // Choose its parent else keep root // Make an entry for all those email address mapping[a[i]] = root;}
}// Create the necessary structure for result 
unordered_map<string, set<string>> mp;
for (auto &p : mapping) mp[p.second].insert(p.first);// Form the resultant vector
for (auto &pp : mp) {vector<string> emails(pp.second.begin(), pp.second.end());emails.insert(emails.begin(), pp.first);result.push_back(emails);
}return result;

题解二:

Bad Performance Solution

//Bad Performance Solution
class Solution_Time_Limit_Exceeded {
public:// We can orginze all relevant emails to a chain,// then we can use Union Find algorithm// Besides, we also need to map the relationship between name and email.vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {unordered_map<string, string> emails_chains; // email chainsunordered_map<string, string> names; // names to email chains' head//initializationfor(int i = 0 ; i<accounts.size();i++) {auto& account = accounts[i];auto& name = account[0];for (int j=1; j<account.size(); j++) {auto& email = account[j];if ( names.find(email) == names.end() ) {emails_chains[email] = email;names[email] = name;}join(emails_chains, account[1], email);}}//reform the emailsunordered_map<string, set<string>> res;for( auto& acc : accounts ) {string e = find(emails_chains, acc[1]);res[e].insert(acc.begin()+1, acc.end());}//output the resultvector<vector<string>> result;for (auto pair : res) {vector<string> emails(pair.second.begin(), pair.second.end());emails.insert(emails.begin(), names[pair.first]);result.push_back(emails);}return result;}string find(unordered_map<string, string>& emails_chains,string email) {while( email != emails_chains[email] ){email = emails_chains[email];}return email;}bool join(unordered_map<string, string>& emails_chains,string& email1, string& email2) {string e1 = find(emails_chains, email1);string e2 = find(emails_chains, email2);if ( e1 != e2 )  emails_chains[e1] = email2;return e1 == e2;}
};

题解三:

Performance Tunning

//
// Performance Tunning
// -----------------
//
// The above algorithm need to do string comparison, it causes lots of efforts
// So, we allocated the ID for each email, and compare the ID would save the time.
//
// Furthermore, we can use the Group-Email-ID instead of email ID,
// this would save more time.
//
class Solution {
public:// We can orginze all relevant emails to a chain,// then we can use Union Find algorithm// Besides, we also need to map the relationship between name and email.vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {unordered_map<string, int> emails_id; //using email ID for union findunordered_map<int, int> emails_chains; // email chainsunordered_map<int, string> names; // email id & name//initialization & joinfor(int i = 0 ; i<accounts.size();i++) {// using the account index as the emails group ID,// this could simplify the emails chain.emails_chains[i] = i;auto& account = accounts[i];auto& name = account[0];for (int j=1; j<account.size(); j++) {auto& email = account[j];if ( emails_id.find(email) == emails_id.end() ) {emails_id[email] = i;names[i] = name;}else {join( emails_chains, i, emails_id[email] );}}}//reform the emailsunordered_map<int, set<string>> res;for(int i=0; i<accounts.size(); i++) {int idx = find(emails_chains, i);res[idx].insert(accounts[i].begin()+1, accounts[i].end());}//output the resultvector<vector<string>> result;for (auto pair : res) {vector<string> emails( pair.second.begin(), pair.second.end() );emails.insert(emails.begin(), names[pair.first]);result.push_back(emails);}return result;}int find(unordered_map<int, int>& emails_chains, int id) {while( id != emails_chains[id] ){id = emails_chains[id];}return id;}bool join(unordered_map<int, int>& emails_chains, int id1, int id2) {int e1 = find(emails_chains, id1);int e2 = find(emails_chains, id2);if ( e1 != e2 )  emails_chains[e1] = e2;return e1 == e2;}
};