問題
In C#, an IComparer<T> is an interface: any class that implements its Compare method can tell a sort which item comes first. Choosing a different comparer changes the order without changing the sort.
Sort the people by key:
name: by name (ordinal character order, so uppercase letters come before lowercase), and by age when names are equal.age: by age, youngest first, and by name when ages are equal.
Implement sort_by_key(names, ages, key).
Input
names, ages: The people (same length).
key: "name" or "age".
Output
Return the names in sorted order.
制約
0 ≤ number of people ≤ 100,000
0 ≤ age ≤ 150
例
入力：
key = "age"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
期待される出力：
expected = ["Bo", "Lin", "Ada"]入力：
std::string key = "age";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
期待される出力：
std::vector<std::string> expected = {"Bo", "Lin", "Ada"};入力：
string key = "age";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
期待される出力：
string[] expected = new string[] {"Bo", "Lin", "Ada"};解説
Bo and Lin are both 30, so they are ordered by name before Ada (36): ["Bo", "Lin", "Ada"].
入力：
key = "name"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
期待される出力：
expected = ["Ada", "Bo", "Lin"]入力：
std::string key = "name";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
期待される出力：
std::vector<std::string> expected = {"Ada", "Bo", "Lin"};入力：
string key = "name";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
期待される出力：
string[] expected = new string[] {"Ada", "Bo", "Lin"};解説
Alphabetical order: ["Ada", "Bo", "Lin"].
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
Sort pairs of (name, age) instead of two separate lists.
ヒント 2
The chosen key gives the main order; the other field breaks ties.
ヒント 3
Use ordinal string comparison so uppercase sorts before lowercase.
目安の計算量
O(n log n) time, O(n) 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。