本科只用过C++和Python刷题，原因是感觉Java语言太冗长厚重。 最近尝试了下用Java刷题，总结一下输入输出

最基本的输入输出：
Scanner sc = new Scanner(System.in);//输入 util包下
String next = sc.next();
int x = sc.nextInt();

System.out.println();// 输出
System.out.print();

更快一些的输入输出：
注意引入 io包和 throws IOException
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));//输入
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));//输出
String[] firstline = br.readLine().split(" ");//注意是读取一行
int n=Integer.parseInt(firstline[0]);//转为整型

bw.newLine();//换行
bw.write(1+"");//输出也只能输出字符串，如果需要输出一个数字，则需要+""以转成字符串。

br.close();
bw.flush();// 刷新（需要刷新才能显示）
bw.close();

举例子： 154. 滑动窗口 https://www.acwing.com/problem/content/description/156/

标准输入输出

import java.util.*;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int[] nums = new int[n];
        for(int i = 0; i < n; i++){
            nums[i] = sc.nextInt();
        }
        Deque<Integer> q = new LinkedList<>();
        for(int i = 0; i < n; i++){
            if(q.size() > 0 && i - q.peekFirst() >= k){
                q.pollFirst();
            }
            while(q.size() > 0 && nums[i] <= nums[q.peekLast()]){
                q.pollLast();
            }
            q.add(i);
            if(i >= k - 1){
                System.out.print(nums[q.peekFirst()] + " ");
            }
        }
        System.out.println();
        q.clear();
        for(int i = 0; i < n; i++){
            if(q.size() > 0 && i - q.peekFirst() >= k){
                q.pollFirst();
            }
            while(q.size() > 0 && nums[i] >= nums[q.peekLast()]){
                q.pollLast();
            }
            q.add(i);
            if(i >= k - 1){
                System.out.print(nums[q.peekFirst()] + " ");
            }
        }
    }
}

更快一些的输入输出：

import java.util.*;
import java.io.*;
public class Main{
    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        String[] s0 = br.readLine().split(" ");
        int n = Integer.parseInt(s0[0]);
        int k = Integer.parseInt(s0[1]);

        int[] nums = new int[n];
        String[] s1 = br.readLine().split(" ");
        for(int i = 0; i < n; i++){
            nums[i] = Integer.parseInt(s1[i]);
        }
        Deque<Integer> q = new LinkedList<>();
        for(int i = 0; i < n; i++){
            if(q.size() > 0 && i - q.peekFirst() >= k){
                q.pollFirst();
            }
            while(q.size() > 0 && nums[i] <= nums[q.peekLast()]){
                q.pollLast();
            }
            q.add(i);
            if(i >= k - 1){
                bw.write(nums[q.peekFirst()] + " ");
            }
        }
        bw.newLine();
        q.clear();
        for(int i = 0; i < n; i++){
            if(q.size() > 0 && i - q.peekFirst() >= k){
                q.pollFirst();
            }
            while(q.size() > 0 && nums[i] >= nums[q.peekLast()]){
                q.pollLast();
            }
            q.add(i);
            if(i >= k - 1){
                bw.write(nums[q.peekFirst()] + " ");
            }
        }
        bw.flush();
        bw.close();
        br.close();
    }
}

Java.util.Scanner类是一个简单的文本扫描类，它可以解析基本数据类型和字符串，它本质上其实是使用正则表达式去读取不同的数据类型

Java.io.BufferedReader类为了能够高效的读取字符序列，从字符输入流和字符缓冲区读取文本

BufferedReader的缓冲区大小为8KB，Scanner的缓冲区大小为1KB

Scanner的平均耗时是BufferedReader的10倍左右