并查集（方法一）

class Solution {unordered_map<string, int> index; // 每个邮箱都有一个唯一编号int root[10010];                  // 并查集void unionset(int a, int b) {int rt1 = findrt(a), rt2 = findrt(b);if (rt1 == rt2) return; // 很重要root[rt1] += root[rt2];root[rt2] = rt1;}int findrt(int a) {if (root[a] < 0) return a;return root[a] = findrt(root[a]);}public:vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {memset(root, -1, sizeof(root));// 给出indexfor (int i = 0; i < accounts.size(); i++) {for (int j = 1; j < accounts[i].size(); j++) {index[accounts[i][j]]++;}}int email_cnt = 0;for (auto it = index.begin(); it != index.end(); it++) {it->second = email_cnt++;}// 并for (int i = 0; i < accounts.size(); i++) {for (int j = 2; j < accounts[i].size(); j++) {unionset(index[accounts[i][j - 1]], index[accounts[i][j]]);}}// 查unordered_map<int, int> rt_and_ansIndex;vector<vector<string>> ans;for (int i = 0; i < accounts.size(); i++) {int rt_index = findrt(index[accounts[i][1]]);if (rt_and_ansIndex.count(rt_index) == 0) {rt_and_ansIndex[rt_index] = ans.size();ans.push_back(vector<string>());ans.back().push_back(accounts[i][0]); // 账户名}for (int j = 1; j < accounts[i].size(); j++) {ans[rt_and_ansIndex[rt_index]].push_back(accounts[i][j]); // 邮箱}}for (int i = 0; i < ans.size(); i++) {sort(ans[i].begin() + 1, ans[i].end()); // 排序ans[i].erase(unique(ans[i].begin() + 1, ans[i].end()), ans[i].end()); // 去重}return ans;}
};

dfs（方法二）

class Solution {unordered_map<string, int> index; // 每个邮箱都有一个唯一编号string emails[10010];vector<vector<int>> g; // 图，邻接表存法bool visited[10010];vector<vector<string>> ans;void dfs(int n) {visited[n] = true;ans.back().push_back(emails[n]);for (int i = 0; i < g[n].size(); i++) {if (!visited[g[n][i]]) {dfs(g[n][i]);}}return;}public:vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {memset(visited, false, sizeof(visited));// 给出indexfor (int i = 0; i < accounts.size(); i++) {for (int j = 1; j < accounts[i].size(); j++) {index[accounts[i][j]]++;}}int email_cnt = 0;for (auto it = index.begin(); it != index.end(); it++) {it->second = email_cnt++;emails[it->second] = it->first;}g.resize(email_cnt);// 建图for (int i = 0; i < accounts.size(); i++) {for (int j = 2; j < accounts[i].size(); j++) {g[index[accounts[i][1]]].push_back(index[accounts[i][j]]);g[index[accounts[i][j]]].push_back(index[accounts[i][1]]); //双向边}}// dfsfor (int i = 0; i < accounts.size(); i++) {if (!visited[index[accounts[i][1]]]) {ans.push_back(vector<string>());ans.back().push_back(accounts[i][0]);dfs(index[accounts[i][1]]);}}for (int i = 0; i < ans.size(); i++) {sort(ans[i].begin() + 1, ans[i].end()); // 排序ans[i].erase(unique(ans[i].begin() + 1, ans[i].end()),ans[i].end()); // 去重}return ans;}
};

dfs换一种写法

class Solution {unordered_map<string, vector<string>> g; // 图，邻接表存法,存邮箱，而不是set<string> visited;vector<vector<string>> ans;void dfs(string str) {visited.insert(str);ans.back().push_back(str);for (int i = 0; i < g[str].size(); i++) {if (visited.count(g[str][i]) == 0) {dfs(g[str][i]);}}return;}public:vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {// 建图for (int i = 0; i < accounts.size(); i++) {for (int j = 2; j < accounts[i].size(); j++) {g[accounts[i][1]].push_back(accounts[i][j]);g[accounts[i][j]].push_back(accounts[i][1]);}}// dfsfor (int i = 0; i < accounts.size(); i++) {if (visited.count(accounts[i][1]) == 0) {ans.push_back(vector<string>());ans.back().push_back(accounts[i][0]);dfs(accounts[i][1]);}}for (int i = 0; i < ans.size(); i++) {sort(ans[i].begin() + 1, ans[i].end()); // 排序ans[i].erase(unique(ans[i].begin() + 1, ans[i].end()), ans[i].end()); // 去重}return ans;}
};

如果unordered_map改为map，更慢。