Accounts Merge

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", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'],  ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
Explanation: 
The first and third John's are the same person as they have the common email "johnsmith@mail.com".
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', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], 
['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] 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].

思路：这题作为graph的题目来做，email变成node，name之后可以建立一个图，name单独store出来，然后用email node来做dfs，收集在一条path上面的email，然后最后construct result。 这题BFS收集和DFS收集，都可以； 

class Solution {
    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        List<List<String>> lists = new ArrayList<List<String>>();
        //      email   name
        HashMap<String, String> emailToNameMap = new HashMap<>();
        HashMap<String, HashSet<String>> graph = new HashMap<>();
        for(List<String> account: accounts) {
            String name = account.get(0);
            for(int i = 1; i < account.size(); i++) {
                String email = account.get(i);
                emailToNameMap.put(email, name);
                graph.putIfAbsent(email, new HashSet<>());
                if(i > 1) {
                    graph.get(account.get(i)).add(account.get(i - 1));
                    graph.get(account.get(i - 1)).add(account.get(i));
                }
            }
        }
        
        HashSet<String> visited = new HashSet<>();
        for(String email: graph.keySet()) {
            if(!visited.contains(email)) {
                List<String> list = new LinkedList<>();
                // bfs(graph, email, list, visited);
                dfs(graph, email, list, visited);
                Collections.sort(list); // 别忘记sort；
                list.add(0, emailToNameMap.get(list.get(0)));
                lists.add(list);
            }
        }
        return lists;
    }
    
    private void dfs(HashMap<String, HashSet<String>> graph, String email, List<String> list,
                    HashSet<String> visited) {
        if(!visited.contains(email)) {
            visited.add(email);
            list.add(email);
            for(String neighbor: graph.get(email)) {
                dfs(graph, neighbor, list, visited);
            }
        }
    }
    
    private void bfs(HashMap<String, HashSet<String>> graph, String email, List<String> list,
                    HashSet<String> visited) {
        Queue<String> queue = new LinkedList<>();
        queue.offer(email);
        visited.add(email);
        
        while(!queue.isEmpty()) {
            String node = queue.poll();
            list.add(node);
            for(String neighbor: graph.get(node)) {
                if(!visited.contains(neighbor)) {
                    visited.add(neighbor);
                    queue.offer(neighbor);
                }
            }
        }
    }
}

思路2：经典的union find题目。以email为点来进行union，email刚开始的默认index是row index，此时union find的是accounts的index，0，1，2，如果后面email有相同的，则跟row index union。然后再根据index相同来收集emails。最后加入email去找index，找到name，然后combine成result，学会用putIfAbsent.

class Solution {
    private class UnionFind {
        private int[] father;
        private int count;
        public UnionFind(int n) {
            this.father = new int[n + 1];
            for(int i = 0; i <= n; i++) {
                father[i] = i;
            }
            this.count = n;
        }
        
        public int find(int x) {
            int j = x;
            while(father[j] != j) {
                j = father[j];
            }
            // path compression;
            while(x != j) {
                int fx = father[x];
                father[x] = j;
                x = fx;
            }
            
            return j;
        }
        
        public void union(int a, int b) {
            int root_a = find(a);
            int root_b = find(b);
            if(root_a != root_b) {
                father[root_a] = root_b;
                this.count--;
            }
        }
        
        public int getCount() {
            return this.count;
        }
    }
    
    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        List<List<String>> lists = new ArrayList<List<String>>();
        //      email, rowIndex;
        HashMap<String, Integer> emailToIndex = new HashMap<>();
        // 先按照email union row index；
        UnionFind uf = new UnionFind(accounts.size());
        for(int i = 0; i < accounts.size(); i++) {
            List<String> account = accounts.get(i);
            for(int j = 1; j < account.size(); j++) {
                String email = account.get(j);
                emailToIndex.putIfAbsent(email, i);
                uf.union(i, emailToIndex.get(email));
            }
        }
        
        // 按照row index收集root index的所有email
        HashMap<Integer, List<String>> indexToEmails = new HashMap<>();
        for(String email: emailToIndex.keySet()) {
            int index = emailToIndex.get(email);
            int root = uf.find(index);
            indexToEmails.putIfAbsent(root, new ArrayList<String>());
            indexToEmails.get(root).add(email);
        }
        
        // 收集之后，sort，然后加入name；
        for(Integer integer: indexToEmails.keySet()) {
            List<String> list = new ArrayList<>();
            list.addAll(indexToEmails.get(integer));
            Collections.sort(list);
            list.add(0, accounts.get(integer).get(0));
            lists.add(list);
        }
        return lists;
    }
}