Problème
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.
Contraintes
0 ≤ number of people ≤ 100,000
0 ≤ age ≤ 150
Exemples
Entrée :
key = "age"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
Sortie attendue :
expected = ["Bo", "Lin", "Ada"]Entrée :
std::string key = "age";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
Sortie attendue :
std::vector<std::string> expected = {"Bo", "Lin", "Ada"};Entrée :
string key = "age";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
Sortie attendue :
string[] expected = new string[] {"Bo", "Lin", "Ada"};Explication
Bo and Lin are both 30, so they are ordered by name before Ada (36): ["Bo", "Lin", "Ada"].
Entrée :
key = "name"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
Sortie attendue :
expected = ["Ada", "Bo", "Lin"]Entrée :
std::string key = "name";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
Sortie attendue :
std::vector<std::string> expected = {"Ada", "Bo", "Lin"};Entrée :
string key = "name";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
Sortie attendue :
string[] expected = new string[] {"Ada", "Bo", "Lin"};Explication
Alphabetical order: ["Ada", "Bo", "Lin"].
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
Sort pairs of (name, age) instead of two separate lists.
Indice 2
The chosen key gives the main order; the other field breaks ties.
Indice 3
Use ordinal string comparison so uppercase sorts before lowercase.
Complexité attendue
O(n log n) time, O(n) extra space
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.