小凯的排序课

本题为签到题，考察简单 for 循环和 if-else if-else 语句。此外，可以使用映射关系的数据结构，来减少代码量，提高编码效率

C++ 实现

#include <bits/stdc++.h>

using namespace std;
unordered_map<string, string> dict = {
    {"quick sort"    , "O(nlogn) O(logn) No" },
    {"merge sort"    , "O(nlogn) O(n) Yes"   },
    {"heap sort"     , "O(nlogn) O(1) No"    },
    {"shell sort"    , "O(n^{1.3~2}) O(1) No"},
    {"counting sort" , "O(n+m) O(n+m) Yes"   },
    {"insertion sort", "O(n^2) O(1) Yes"     },
    {"bubble sort"   , "O(n^2) O(1) Yes"     },
    {"selection sort", "O(n^2) O(1) No"      },
};
void solve() {
    string s;
    getline(cin, s);
    cout << dict[s] << endl;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    cin.ignore();    // 注意要 ignore，否则第一行的行尾换行符会影响 getline
    while (t--)
        solve();
    return 0;
}

Python 实现

mp = {
    "quick sort": "O(nlogn) O(logn) No",
    "merge sort": "O(nlogn) O(n) Yes",
    "heap sort": "O(nlogn) O(1) No",
    "shell sort": "O(n^{1.3~2}) O(1) No",
    "counting sort": "O(n+m) O(n+m) Yes",
    "insertion sort": "O(n^2) O(1) Yes",
    "bubble sort": "O(n^2) O(1) Yes",
    "selection sort": "O(n^2) O(1) No",
}
for _ in range(int(input())):
    print(mp[input()])

Java 实现

import java.util.*;

public class Main {
    private static final Scanner scanner = new Scanner(System.in);
    private static final HashMap dict = new HashMap() {{
        put("quick sort"    , "O(nlogn) O(logn) No" );
        put("merge sort"    , "O(nlogn) O(n) Yes"   );
        put("heap sort"     , "O(nlogn) O(1) No"    );
        put("shell sort"    , "O(n^{1.3~2}) O(1) No");
        put("counting sort" , "O(n+m) O(n+m) Yes"   );
        put("insertion sort", "O(n^2) O(1) Yes"     );
        put("bubble sort"   , "O(n^2) O(1) Yes"     );
        put("selection sort", "O(n^2) O(1) No"      );
    }};

    private static void solve() {
        System.out.println(dict.get(scanner.nextLine()));
    }

    public static void main(String[] args) {
        int t = scanner.nextInt();
        for (scanner.skip("\n"); t != 0; t--)   // skip同理C++的ignore
            solve();
        scanner.close();
    }
}