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
class ListNode { int val; ListNode next; ListNode() {} ListNode(int val) { this.val = val; } ListNode(int val, ListNode next) { this.val = val; this.next = next; } } public class Main { public static void main(String[] args) { ListNode l1 = new ListNode(-1); // -1->0->1->2; l1.next = new ListNode(0); l1.next.next = new ListNode(1); l1.next.next.next = new ListNode(2); ListNode l2 = new ListNode(-1);// -1 -> -2 -> ->1->2; l2.next = new ListNode(-2); // l2.next.next = l1; // 让 l2 的最后一个节点指向 l1 的第二个节点 ListNode pa = l1, pb = l2; while (pa != pb) { pa = pa == null ? l2 : pa.next; pb = pb == null ? l1 : pb.next; } if (pb != null) { System.out.print(pb.val); } else { System.out.print("No intersection"); } } }import java.util.*; public class Main { public static int N = 1010; public static int[] f = new int[N]; // 存储以 i 结尾的最长上升子序列长度 public static int[] a = new int[N]; // 存储输入的序列 public static int[] pre = new int[N]; // 存储每个位置的前驱元素 public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); for (int i = 1; i <= n; i++) a[i] = sc.nextInt(); // 初始化 for (int i = 1; i <= n; i++) { f[i] = 1; pre[i] = -1; // 初始化前驱为 -1，表示没有前驱 } // 动态规划求解最长上升子序列 for (int i = 1; i <= n; i++) { for (int j = 1; j < i; j++) { if (a[i] > a[j] && f[i] < f[j] + 1) { f[i] = f[j] + 1; pre[i] = j; // 记录前驱 } } } // 找到最长上升子序列的最后一个元素的位置 int res = 0; int end = 0; for (int i = 1; i <= n; i++) { if (f[i] > res) { res = f[i]; end = i; } } // 输出最长上升子序列的长度 System.out.println(res); // 输出最长上升子序列 int[] list = new int[res]; int idx = res - 1; while (end != -1) { list[idx--] = a[end]; end = pre[end]; } for (int i = 0; i < res; i++) { System.out.print(lis[i] + " "); } } }while ((line = in.readLine())!= null) parts = line.split(" ");