给定一个列表 accounts，每个元素 accounts[i] 是一个字符串列表，其中第一个元素 accounts[i][0] 是 名称 (name)，其余元素是 emails 表示该账户的邮箱地址。
现在，我们想合并这些账户。如果两个账户都有一些共同的邮箱地址，则两个账户必定属于同一个人。请注意，即使两个账户具有相同的名称，它们也可能属于不同的人，因为人们可能具有相同的名称。一个人最初可以拥有任意数量的账户，但其所有账户都具有相同的名称。
合并账户后，按以下格式返回账户：每个账户的第一个元素是名称，其余元素是 按字符 ASCII 顺序排列 的邮箱地址。账户本身可以以 任意顺序 返回。
示例 1：
输入：accounts = [["John", "[email protected]", "[email protected]"], ["John", "[email protected]"], ["John", "[email protected]", "[email protected]"], ["Mary", "[email protected]"]] 输出：[["John", '[email protected]', '[email protected]', '[email protected]'], ["John", "[email protected]"], ["Mary", "[email protected]"]] 解释： 第一个和第三个 John 是同一个人，因为他们有共同的邮箱地址 "[email protected]"。 第二个 John 和 Mary 是不同的人，因为他们的邮箱地址没有被其他帐户使用。 可以以任何顺序返回这些列表，例如答案 [['Mary'，'[email protected]']，['John'，'[email protected]']， ['John'，'[email protected]'，'[email protected]'，'[email protected]']] 也是正确的。
示例 2：
输入：accounts = [["Gabe","[email protected]","[email protected]","[email protected]"],["Kevin","[email protected]","[email protected]","[email protected]"],["Ethan","[email protected]","[email protected]","[email protected]"],["Hanzo","[email protected]","[email protected]","[email protected]"],["Fern","[email protected]","[email protected]","[email protected]"]] 输出：[["Ethan","[email protected]","[email protected]","[email protected]"],["Gabe","[email protected]","[email protected]","[email protected]"],["Hanzo","[email protected]","[email protected]","[email protected]"],["Kevin","[email protected]","[email protected]","[email protected]"],["Fern","[email protected]","[email protected]","[email protected]"]]
提示：
1 <= accounts.length <= 10002 <= accounts[i].length <= 101 <= accounts[i][j].length <= 30accounts[i][0]由英文字母组成accounts[i][j] (for j > 0)是有效的邮箱地址
并查集。
模板 1——朴素并查集：
# 初始化，p存储每个点的父节点
p = list(range(n))
# 返回x的祖宗节点
def find(x):
if p[x] != x:
# 路径压缩
p[x] = find(p[x])
return p[x]
# 合并a和b所在的两个集合
p[find(a)] = find(b)模板 2——维护 size 的并查集：
# 初始化，p存储每个点的父节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
p = list(range(n))
size = [1] * n
# 返回x的祖宗节点
def find(x):
if p[x] != x:
# 路径压缩
p[x] = find(p[x])
return p[x]
# 合并a和b所在的两个集合
if find(a) != find(b):
size[find(b)] += size[find(a)]
p[find(a)] = find(b)模板 3——维护到祖宗节点距离的并查集：
# 初始化，p存储每个点的父节点，d[x]存储x到p[x]的距离
p = list(range(n))
d = [0] * n
# 返回x的祖宗节点
def find(x):
if p[x] != x:
t = find(p[x])
d[x] += d[p[x]]
p[x] = t
return p[x]
# 合并a和b所在的两个集合
p[find(a)] = find(b)
d[find(a)] = distance对于本题，初始每个 account 是一个独立的集合。遍历 accounts 列表，若遇到相同的 email，则将两账户进行合并，遍历完毕得到并查集。
接着对每个集合进行中的所有账户邮箱进行合并排序，最后返回账户名称以及对应的邮箱列表。
class Solution:
def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
n = len(accounts)
p = list(range(n))
email_id = {}
for i, account in enumerate(accounts):
name = account[0]
for email in account[1:]:
if email in email_id:
p[find(i)] = find(email_id[email])
else:
email_id[email] = i
mp = defaultdict(set)
for i, account in enumerate(accounts):
for email in account[1:]:
mp[find(i)].add(email)
ans = []
for i, emails in mp.items():
t = [accounts[i][0]]
t.extend(sorted(emails))
ans.append(t)
return ansclass Solution {
private int[] p;
public List<List<String>> accountsMerge(List<List<String>> accounts) {
int n = accounts.size();
p = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
}
Map<String, Integer> emailId = new HashMap<>();
for (int i = 0; i < n; ++i) {
List<String> account = accounts.get(i);
String name = account.get(0);
for (int j = 1; j < account.size(); ++j) {
String email = account.get(j);
if (emailId.containsKey(email)) {
p[find(i)] = find(emailId.get(email));
} else {
emailId.put(email, i);
}
}
}
Map<Integer, Set<String>> mp = new HashMap<>();
for (int i = 0; i < n; ++i) {
List<String> account = accounts.get(i);
for (int j = 1; j < account.size(); ++j) {
String email = account.get(j);
mp.computeIfAbsent(find(i), k -> new HashSet<>()).add(email);
}
}
List<List<String>> res = new ArrayList<>();
for (Map.Entry<Integer, Set<String>> entry : mp.entrySet()) {
List<String> t = new LinkedList<>();
t.addAll(entry.getValue());
Collections.sort(t);
t.add(0, accounts.get(entry.getKey()).get(0));
res.add(t);
}
return res;
}
private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}class Solution {
public:
vector<int> p;
vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
int n = accounts.size();
p.resize(n);
for (int i = 0; i < n; ++i) p[i] = i;
unordered_map<string, int> emailId;
for (int i = 0; i < n; ++i) {
auto account = accounts[i];
auto name = account[0];
for (int j = 1; j < account.size(); ++j) {
string email = account[j];
if (emailId.count(email))
p[find(i)] = find(emailId[email]);
else
emailId[email] = i;
}
}
unordered_map<int, unordered_set<string>> mp;
for (int i = 0; i < n; ++i) {
auto account = accounts[i];
for (int j = 1; j < account.size(); ++j) {
string email = account[j];
mp[find(i)].insert(email);
}
}
vector<vector<string>> ans;
for (auto& [i, emails] : mp) {
vector<string> t;
t.push_back(accounts[i][0]);
for (string email : emails) t.push_back(email);
sort(t.begin() + 1, t.end());
ans.push_back(t);
}
return ans;
}
int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
};