[BOJ][10816] 숫자 카드 2

1. 문제 링크

https://www.acmicpc.net/problem/10816

2. 사용 알고리즘

구현

3. 풀이

배열에 카드 별 개수를 저장한 후 저장된 개수 출력

4. 소스 코드

4-1. C++

https://github.com/dev-aiden/problem-solving/blob/main/boj/10816.cpp

#include <iostream>
#include <algorithm>

using namespace std;

int n, m, idx;
int arr[500003], arr2[500003];
int cnt[20000003];

int main(void) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    cin >> n;
    for(int i = 0; i < n; ++i) {
        int num; cin >> num;
        ++cnt[num + 10000000];
    }
    cin >> m;
    for(int i = 0; i < m; ++i) cin >> arr2[i];
    for(int i = 0; i < m; ++i) cout << cnt[arr2[i] + 10000000] << " ";
    cout << "\n";
    return 0;
}

4-2. JAVA

https://github.com/dev-aiden/problem-solving/blob/main/boj/10816.java

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {

    static int n, m, idx;
    static int[] arr = new int[500003];
    static int[] arr2 = new int[500003];
    static int[] cnt = new int[20000003];

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        n = Integer.parseInt(br.readLine());
        StringTokenizer st = new StringTokenizer(br.readLine());
        for(int i = 0; i < n; ++i) {
            int num = Integer.parseInt(st.nextToken());
            ++cnt[num + 10000000];
        }
        m = Integer.parseInt(br.readLine());
        st = new StringTokenizer(br.readLine());
        for(int i = 0; i < m; ++i) arr2[i] = Integer.parseInt(st.nextToken());
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < m; ++i) sb.append(cnt[arr2[i] + 10000000]).append(" ");
        sb.append("\n");
        System.out.println(sb);
    }
}