1169. Invalid Transactions

问题：

给定由【姓名，时间，数额，城市】组成的交易信息数组。

求无效交易的数组。

无效：1 or 2满足

1，数额>1000

2，相同姓名，不同城市，时间差<60

Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.

Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]

Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
 

Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000.

　　

解法：

姓名相同，使用unordered_map存储，以姓名为索引key，

value为 剩下的3个情报的，map，方便查找。

每读入一条记录，在已经存储的map中寻找到，和本条记录互为 无效交易的时候，

插入结果set，防止重复，因此使用set

⚠️ 注意：读入记录，并进行分割时，使用输入流 istringstream

使用getline方法，根据分隔符 ','  来循环取得各个区域内容。

for(string& sac:transactions){
//对每一条记录
            istringstream ss(sac);
            vector<string> s(4,"");
            int i=0;
            while(getline(ss, s[i++], ','));
}

代码参考：

class Solution {
public:
    vector<string> invalidTransactions(vector<string>& transactions) {
        unordered_map<string, vector<vector<string>>> transct;
        set<string> res;
        for(string& sac:transactions){
            istringstream ss(sac);
            vector<string> s(4,"");
            int i=0;
            while(getline(ss, s[i++], ','));
            
            if(stoi(s[2])>1000) res.insert(sac);//amount
            
            for(vector<string> p:transct[s[0]]){//same name: time,amount,city
                if(abs(stoi(p[0])-stoi(s[1]))<=60 && p[2]!=s[3]){
                    res.insert(s[0]+","+p[0]+","+p[1]+","+p[2]);
                    if(res.count(sac)==0) res.insert(sac);
                }
            }
            transct[s[0]].push_back({s[1],s[2],s[3]});
        }
        vector<string> ans;
        for(string r:res) ans.push_back(r);
        return ans;
    }
};