721. 账户合并
作者:互联网
并查集就能实现,我都忘了并查集了,唉
用map记录每个邮箱的账号,再此之前判断该邮箱是否已经在map中有value,即该邮箱是否已经标记了账户
如果有,那就将当前邮箱的所有账号,转移到标记的账户
如果当前账户下的多个邮箱在map中有不同的value,那就统一转移到第一个value即可
用set存储每个邮箱的账户
class Solution { public: set<string> s[1010]; vector<vector<string>> ret; map<string, int> mmap; int vis[1010]; void move(int flag, int k) { flag -= 1; k -= 1; vis[k] = 0; for(set<string>::iterator it = s[k].begin(); it != s[k].end(); it++) { s[flag].insert(*it); mmap[*it] = flag + 1; } } vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) { int len1 = accounts.size(); memset(vis, 0, sizeof(vis)); for(int i = 0; i < len1; i++) { int len2 = accounts[i].size(); int flag = -1; for(int j = 1; j < len2; j++) { if(mmap[accounts[i][j]]) { if(flag == -1) flag = mmap[accounts[i][j]]; else if(mmap[accounts[i][j]] != flag) move(flag, mmap[accounts[i][j]]); } } if(flag == -1) { vis[i] = 1; for(int j = 1; j < len2; j++) { s[i].insert(accounts[i][j]); mmap[accounts[i][j]] = i + 1; } } else { for(int j = 1; j < len2; j++) { s[flag - 1].insert(accounts[i][j]); mmap[accounts[i][j]] = flag; } } } int ans = 0; vector<string> v; for(int i = 0; i < len1; i++) { v.clear(); if(vis[i]) { // cout << 111 << endl; v.push_back(accounts[i][0]); for(set<string>::iterator it = s[i].begin(); it != s[i].end(); it++) { v.push_back(*it); } ret.push_back(v); } } return ret; } };
标签:vis,账户,int,mmap,合并,++,flag,accounts,721 来源: https://www.cnblogs.com/WTSRUVF/p/15381327.html