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时限超过了最佳方法(Java)

java

Csa77 Bartosz RadaczyÅski · 技术社区 · 6 年前

即使我认为我解决了一个有竞争力的程序 problem from HackerEarth 使用最佳方法,所有测试都会超过时间限制。我真的不知道如何进一步优化它,因为这只是一个简单的练习。

我的方法:迭代所有数组成员,然后将它们添加到 HashMap 存储它们的事件。之后,只需读取查询编号并从 哈希图 .

这是我的解决方案:

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;

class TestClass {

    public static void main(String args[]) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());

        //go through all test cases
        for (int i = 0; i < t; i++) {
            Map<Integer, Integer> map = new HashMap<>();
            String[] inputs = br.readLine().split(" ");
            int N = Integer.parseInt(inputs[0]);
            int Q = Integer.parseInt(inputs[1]);
            inputs = br.readLine().split(" ");

            //read array
            for (int j = 0; j < N; j++) {
                int x = Integer.parseInt(inputs[j]);
                Integer value = map.get(x);
                //if number is already in hashmap then increment its count
                //else put it into the map with a count of 1
                if (value == null) {
                    map.put(x, 1);
                } else map.put(x, value + 1);
            }

            //iterate through the queries and get their occurences from the map
            for (int j = 0; j < Q; j++) {
                int x = Integer.parseInt(br.readLine());
                Integer value = map.get(x);
                if (value == null) {
                    System.out.println(0);
                } else System.out.println(value);
            }
        }
    }
}

我的问题是:我的方法有什么问题?为什么时间用完了?

2 回复 | 直到 6 年前

tomky71 6 年前

PrinteWriter

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.Map;

class TestClass {

    public static void main(String args[]) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pr = new PrintWriter(System.out);
        int t = Integer.parseInt(br.readLine());

        //go through all test cases
        for (int i = 0; i < t; i++) {
            Map<Integer, Integer> map = new HashMap<>();
            String[] inputs = br.readLine().split(" ");
            int N = Integer.parseInt(inputs[0]);
            int Q = Integer.parseInt(inputs[1]);
            inputs = br.readLine().split(" ");

            //read array
            for (int j = 0; j < N; j++) {
                int x = Integer.parseInt(inputs[j]);
                Integer value = map.get(x);
                //if number is already in hashmap then increment its count
                //else put it into the map with a count of 1
                if (value == null) {
                    map.put(x, 1);
                } else map.put(x, value + 1);
            }

            //iterate through the queries and get their occurences from the map
            for (int j = 0; j < Q; j++) {
                int x = Integer.parseInt(br.readLine());
                Integer value = map.get(x);
                if (value == null) {
                    pr.println(0);
                } else pr.println(value);
            }
        }
        pr.close();
    }
}

Philip DiSarro 6 年前

import java.util.*;
class TestClass {
    public static void main(String args[] ) throws Exception {
        Scanner s = new Scanner(System.in);
        int T = s.nextInt();       
        Map<Integer,Integer> map=new HashMap<Integer,Integer>();
        for(int i=0;i<T;i++)
        {
            StringBuilder sb=new StringBuilder();
            int N=s.nextInt();
            int Q=s.nextInt();
            int[] arr=new int[N];
            for(int j=0;j<N;j++)
            {
                arr[j]=s.nextInt();
                if(map.containsKey(arr[j]))
                {
                    map.put(arr[j],map.get(arr[j])+1);
                }
                else
                map.put(arr[j],1);
            }
            for(int k=0;k<Q;k++)
            {
                int X=s.nextInt();
                if(map.containsKey(X)){
                    sb.append(map.get(X)+"\n");
                }
                else{
                    sb.append(0+"\n");
                }     
            }
             System.out.println(sb.toString()); 
            map.clear();
        }

    }
}

 String[] inputs = br.readLine().split(" ");
 int N = Integer.parseInt(inputs[0]);
 int Q = Integer.parseInt(inputs[1]);
 inputs = br.readLine().split(" ");

Scanner vs. BufferedReader Integer.parseInt(...) br.readlines().split(" ")

Integer value = map.get(x);
if (value == null) {
    pr.println(0);
} else pr.println(value);

map.containsKey(x)

inputs = br.readLine().split(" "); Integer.parseInt(..) s.nextInt() br.readLine() parseInt()