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feat: add solutions to lc problem: No.0721.Accounts Merge
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‎solution/0700-0799/0721.Accounts Merge/README.md

+145-3
Original file line numberDiff line numberDiff line change
@@ -38,27 +38,169 @@ accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnyb
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<li><code>accounts[i][j]</code>的长度将在<code>[1，30]</code>的范围内。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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并查集。
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模板 1——朴素并查集：
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```python
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# 初始化，p存储每个点的父节点
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p = list(range(n))
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# 返回x的祖宗节点
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def find(x):
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if p[x] != x:
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# 路径压缩
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p[x] = find(p[x])
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return p[x]
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# 合并a和b所在的两个集合
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p[find(a)] = find(b)
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```
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模板 2——维护 size 的并查集：
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```python
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# 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
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p = list(range(n))
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size = [1] * n
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# 返回x的祖宗节点
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def find(x):
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if p[x] != x:
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# 路径压缩
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p[x] = find(p[x])
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return p[x]
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# 合并a和b所在的两个集合
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if find(a) != find(b):
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size[find(b)] += size[find(a)]
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p[find(a)] = find(b)
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```
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模板 3——维护到祖宗节点距离的并查集：
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```python
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# 初始化，p存储每个点的父节点，d[x]存储x到p[x]的距离
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p = list(range(n))
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d = [0] * n
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# 返回x的祖宗节点
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def find(x):
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if p[x] != x:
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t = find(p[x])
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d[x] += d[p[x]]
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p[x] = t
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return p[x]
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# 合并a和b所在的两个集合
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p[find(a)] = find(b)
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d[find(a)] = distance
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```
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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。
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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。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
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n = len(accounts)
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p = list(range(n))
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def find(x):
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if p[x] != x:
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p[x] = find(p[x])
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return p[x]
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email_id = {}
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for i in range(n):
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name = accounts[i][0]
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for email in accounts[i][1:]:
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if email in email_id:
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p[find(i)] = find(email_id[email])
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else:
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email_id[email] = i
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mp = collections.defaultdict(set)
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for i in range(n):
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pa = find(i)
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for email in accounts[i][1:]:
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mp[pa].add(email)
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res = []
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for i, emails in mp.items():
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t = [accounts[i][0]]
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t.extend(sorted(emails))
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res.append(t)
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return res
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private int[] p;
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public List<List<String>> accountsMerge(List<List<String>> accounts) {
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int n = accounts.size();
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p = new int[n];
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for (int i = 0; i < n; ++i) {
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p[i] = i;
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}
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Map<String, Integer> emailId = new HashMap<>();
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for (int i = 0; i < n; ++i) {
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List<String> account = accounts.get(i);
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String name = account.get(0);
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for (int j = 1; j < account.size(); ++j) {
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String email = account.get(j);
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if (emailId.containsKey(email)) {
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p[find(i)] = find(emailId.get(email));
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} else {
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emailId.put(email, i);
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}
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}
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}
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Map<Integer, Set<String>> mp = new HashMap<>();
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for (int i = 0; i < n; ++i) {
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int pa = find(i);
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List<String> account = accounts.get(i);
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for (int j = 1; j < account.size(); ++j) {
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String email = account.get(j);
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if (!mp.containsKey(pa)) {
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mp.put(pa, new HashSet<>());
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}
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mp.get(pa).add(email);
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}
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}
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List<List<String>> res = new ArrayList<>();
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for (Map.Entry<Integer, Set<String>> entry : mp.entrySet()) {
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List<String> t = new LinkedList<>();
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t.addAll(entry.getValue());
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Collections.sort(t);
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t.add(0, accounts.get(entry.getKey()).get(0));
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res.add(t);
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}
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return res;
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}
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private int find(int x) {
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if (p[x] != x) {
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p[x] = find(p[x]);
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}
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return p[x];
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}
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}
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```
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### **...**

‎solution/0700-0799/0721.Accounts Merge/README_EN.md

+82-3
Original file line numberDiff line numberDiff line change
@@ -41,21 +41,100 @@ We could return these lists in any order, for example the answer [[&#39;Mary&#39
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<li><code>accounts[i][j] (for j &gt; 0)</code> is a valid email.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
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n = len(accounts)
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p = list(range(n))
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def find(x):
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if p[x] != x:
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p[x] = find(p[x])
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return p[x]
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email_id = {}
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for i in range(n):
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name = accounts[i][0]
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for email in accounts[i][1:]:
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if email in email_id:
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p[find(i)] = find(email_id[email])
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else:
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email_id[email] = i
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mp = collections.defaultdict(set)
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for i in range(n):
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pa = find(i)
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for email in accounts[i][1:]:
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mp[pa].add(email)
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res = []
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for i, emails in mp.items():
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t = [accounts[i][0]]
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t.extend(sorted(emails))
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res.append(t)
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return res
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```
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### **Java**
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```java
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class Solution {
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private int[] p;
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public List<List<String>> accountsMerge(List<List<String>> accounts) {
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int n = accounts.size();
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p = new int[n];
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for (int i = 0; i < n; ++i) {
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p[i] = i;
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}
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Map<String, Integer> emailId = new HashMap<>();
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for (int i = 0; i < n; ++i) {
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List<String> account = accounts.get(i);
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String name = account.get(0);
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for (int j = 1; j < account.size(); ++j) {
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String email = account.get(j);
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if (emailId.containsKey(email)) {
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p[find(i)] = find(emailId.get(email));
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} else {
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emailId.put(email, i);
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}
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}
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}
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Map<Integer, Set<String>> mp = new HashMap<>();
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for (int i = 0; i < n; ++i) {
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int pa = find(i);
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List<String> account = accounts.get(i);
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for (int j = 1; j < account.size(); ++j) {
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String email = account.get(j);
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if (!mp.containsKey(pa)) {
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mp.put(pa, new HashSet<>());
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}
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mp.get(pa).add(email);
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}
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}
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List<List<String>> res = new ArrayList<>();
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for (Map.Entry<Integer, Set<String>> entry : mp.entrySet()) {
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List<String> t = new LinkedList<>();
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t.addAll(entry.getValue());
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Collections.sort(t);
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t.add(0, accounts.get(entry.getKey()).get(0));
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res.add(t);
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}
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return res;
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}
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private int find(int x) {
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if (p[x] != x) {
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p[x] = find(p[x]);
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}
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return p[x];
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}
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}
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```
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### **...**

‎solution/0700-0799/0721.Accounts Merge/Solutioin.java

+52
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,52 @@
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class Solution {
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private int[] p;
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public List<List<String>> accountsMerge(List<List<String>> accounts) {
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int n = accounts.size();
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p = new int[n];
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for (int i = 0; i < n; ++i) {
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p[i] = i;
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}
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Map<String, Integer> emailId = new HashMap<>();
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for (int i = 0; i < n; ++i) {
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List<String> account = accounts.get(i);
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String name = account.get(0);
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for (int j = 1; j < account.size(); ++j) {
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String email = account.get(j);
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if (emailId.containsKey(email)) {
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p[find(i)] = find(emailId.get(email));
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} else {
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emailId.put(email, i);
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}
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}
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}
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Map<Integer, Set<String>> mp = new HashMap<>();
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for (int i = 0; i < n; ++i) {
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int pa = find(i);
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List<String> account = accounts.get(i);
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for (int j = 1; j < account.size(); ++j) {
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String email = account.get(j);
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if (!mp.containsKey(pa)) {
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mp.put(pa, new HashSet<>());
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}
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mp.get(pa).add(email);
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}
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}
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List<List<String>> res = new ArrayList<>();
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for (Map.Entry<Integer, Set<String>> entry : mp.entrySet()) {
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List<String> t = new LinkedList<>();
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t.addAll(entry.getValue());
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Collections.sort(t);
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t.add(0, accounts.get(entry.getKey()).get(0));
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res.add(t);
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}
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return res;
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}
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private int find(int x) {
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if (p[x] != x) {
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p[x] = find(p[x]);
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}
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return p[x];
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}
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}

‎solution/0700-0799/0721.Accounts Merge/Solution.py

+30
Original file line numberDiff line numberDiff line change
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1+
class Solution:
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def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
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n = len(accounts)
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p = list(range(n))
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def find(x):
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if p[x] != x:
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p[x] = find(p[x])
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return p[x]
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email_id = {}
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for i in range(n):
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name = accounts[i][0]
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for email in accounts[i][1:]:
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if email in email_id:
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p[find(i)] = find(email_id[email])
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else:
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email_id[email] = i
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mp = collections.defaultdict(set)
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for i in range(n):
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pa = find(i)
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for email in accounts[i][1:]:
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mp[pa].add(email)
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res = []
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for i, emails in mp.items():
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t = [accounts[i][0]]
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t.extend(sorted(emails))
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res.append(t)
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return res

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