Problem
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.
Einschränkungen
0 ≤ number of people ≤ 100,000
0 ≤ age ≤ 150
Beispiele
Eingabe:
key = "age"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
Erwartete Ausgabe:
expected = ["Bo", "Lin", "Ada"]Eingabe:
std::string key = "age";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
Erwartete Ausgabe:
std::vector<std::string> expected = {"Bo", "Lin", "Ada"};Eingabe:
string key = "age";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
Erwartete Ausgabe:
string[] expected = new string[] {"Bo", "Lin", "Ada"};Erläuterung
Bo and Lin are both 30, so they are ordered by name before Ada (36): ["Bo", "Lin", "Ada"].
Eingabe:
key = "name"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
Erwartete Ausgabe:
expected = ["Ada", "Bo", "Lin"]Eingabe:
std::string key = "name";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
Erwartete Ausgabe:
std::vector<std::string> expected = {"Ada", "Bo", "Lin"};Eingabe:
string key = "name";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
Erwartete Ausgabe:
string[] expected = new string[] {"Ada", "Bo", "Lin"};Erläuterung
Alphabetical order: ["Ada", "Bo", "Lin"].
So gehst du vor
Versuch es zuerst allein und öffne dann einen Hinweis nach dem anderen.
Hinweis 1
Sort pairs of (name, age) instead of two separate lists.
Hinweis 2
The chosen key gives the main order; the other field breaks ties.
Hinweis 3
Use ordinal string comparison so uppercase sorts before lowercase.
Erwartete Komplexität
O(n log n) time, O(n) extra space
Referenzen
Diese Aufgabe wurde für CodeUtility angepasst. Beachte die oben genannten Vorgaben und Einschränkungen.