問題
C# records compare by value: two records are equal when all their fields are equal, even if they are different objects. That makes removing duplicates easy.
Each record is written name,city. Trim spaces around both fields; after trimming, two records are equal when both fields match exactly (case matters).
Implement unique_records(records).
Input
records: The records, in order.
Output
Return the distinct records in the order they first appear, written name,city with the trimmed fields.
制約
0 ≤ number of records ≤ 100,000
Each field has 1 to 30 characters and no commas
例
入力：
records = ["Ada,London", "Lin, Hanoi", "Ada , London"]
期待される出力：
expected = ["Ada,London", "Lin,Hanoi"]入力：
std::vector<std::string> records = {"Ada,London", "Lin, Hanoi", "Ada , London"};
期待される出力：
std::vector<std::string> expected = {"Ada,London", "Lin,Hanoi"};入力：
string[] records = new string[] {"Ada,London", "Lin, Hanoi", "Ada , London"};
期待される出力：
string[] expected = new string[] {"Ada,London", "Lin,Hanoi"};解説
After trimming, the third record equals the first, so it is dropped: ["Ada,London", "Lin,Hanoi"].
入力：
records = ["ada,Paris", "Ada,Paris"]
期待される出力：
expected = ["ada,Paris", "Ada,Paris"]入力：
std::vector<std::string> records = {"ada,Paris", "Ada,Paris"};
期待される出力：
std::vector<std::string> expected = {"ada,Paris", "Ada,Paris"};入力：
string[] records = new string[] {"ada,Paris", "Ada,Paris"};
期待される出力：
string[] expected = new string[] {"ada,Paris", "Ada,Paris"};解説
Case matters, so these two records are different: ["ada,Paris", "Ada,Paris"].
解き方のヒント
まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。
ヒント 1
Trim both fields before comparing records.
ヒント 2
A set of (name, city) pairs remembers which records you have seen.
ヒント 3
Output each new record the first time it appears, in input order.
目安の計算量
O(total length) time, O(number of records) extra space
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。