刷题记录（2）

1. HWOD机试 - 模拟消息队列(100)

package com.yue.test;import org.junit.Test;import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;/*** @Author: 夜雨* @Date: 2021-12-08-10:31* @Description:* @Version 1.0*/
public class tsest01 {@Testpublic void demo01() {String one = "2 22 1 11 4 44 5 55 3 33";String two = "1 7 2 3";LinkedList<String> baseOne = new LinkedList<>(Arrays.asList(one.split(" ")));LinkedList<String> baseTwo = new LinkedList<>(Arrays.asList(two.split(" ")));int resultSize = baseTwo.size() / 2;List<List<Integer>> result = new ArrayList<>();for (int j = baseTwo.size() - 1; j >= 0; j -= 2) {List<Integer> tmp = new ArrayList<>();for (int k = 0; k < baseOne.size() - 1; ) {int time = Integer.parseInt(baseOne.get(k));if (Integer.parseInt(baseTwo.get(j)) > time && Integer.parseInt(baseTwo.get(j - 1)) <= time) {tmp.add(Integer.valueOf(baseOne.get(k + 1)));baseOne.remove(k);baseOne.remove(k);continue;}k += 2;}if (tmp.isEmpty()){tmp.add(-1);}result.add(tmp);}}}

改进

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);String[] aStr = scanner.nextLine().split(" ");int[] a = new int[aStr.length];for (int i = 0; i < aStr.length; i++) {a[i] = Integer.parseInt(aStr[i]);}List<Pair> t = new ArrayList<>();for (int i = 0; i < a.length; i += 2) { // 第一行输入两个一对，为发布者的发布时刻和发布内容t.add(new Pair(a[i], a[i + 1]));}t.sort(null); // 按发布时间排序String[] bStr = scanner.nextLine().split(" ");int[] b = new int[bStr.length];for (int i = 0; i < bStr.length; i++) {b[i] = Integer.parseInt(bStr[i]);}List<List<Integer>> msg = new ArrayList<>();for (int i = 0; i < b.length - 1; i += 2) {msg.add(new ArrayList<>());}for (Pair pair : t) {int x = pair.x;int y = pair.y;int ans = -1;for (int i = b.length - 2; i >= 0; i -= 2) { // 从优先级高的订阅者开始遍历，优先匹配优先级高的订阅者if (b[i] <= x && x < b[i + 1]) { // 如果发布时刻在订阅时间中，订阅者收到发布内容ans = i / 2;break; // 只有优先级最高的收到}}if (ans != -1) {msg.get(ans).add(y); // 收到内容}}for (List<Integer> list : msg) {if (list.isEmpty()) {System.out.println(-1);} else {for (int i : list) {System.out.print(i + " ");}System.out.println();}}}static class Pair implements Comparable<Pair> {int x, y;public Pair(int x, int y) {this.x = x;this.y = y;}@Overridepublic int compareTo(Pair other) {return Integer.compare(this.x, other.x);}}
}

2. 分割数组的最大差值(100)

import java.util.Scanner;
public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();int[] nums = new int[n];long sum = 0;for (int i = 0; i < n; i++) {int num = scanner.nextInt();sum += num;nums[i] = num;}long maxAbs = 0, leftSum = 0;for (int i = 0; i < n - 1; i++) {leftSum += nums[i];// rightSum = sum - leftSum, 差值 = rightSum - leftSum = Math.abs(sum - leftSum)maxAbs = Math.max(maxAbs, Math.abs(sum - 2 * leftSum));}System.out.println(maxAbs);}
}

3. 数值同化(200)-BFS

存在一个m*n的二维数组，其成员取值范围为0,1,2。其中值为1的元素具备同化特性，每经过1s,将上下左右值为0的元素同化为1。而值为2的元素，免疫同化。将数组所有成员随机初始化为0或2，再将矩阵的[0,0]元素修改成1，在经过足够长的时间后，求矩阵中有多少个元素是0或2（即0和2数量之和）
输入：输入的前两个数字是矩阵大小。后面的数字是矩阵内容。
如：4 4
[[0, 0, 0, 0], [0, 2, 2, 2], [0, 2, 0, 0], [0, 2, 0, 0]]
输出：返回矩阵中非1的元素个数

import java.util.*;  public class MatrixAssimilation {  static class Point {  int x, y;  Point(int x, int y) {  this.x = x;  this.y = y;  }  }  public static int countNonOnes(int[][] matrix) {  int m = matrix.length;  int n = matrix[0].length;  // 使用队列来进行BFS  Queue<Point> queue = new LinkedList<>();  // 初始化，将所有初始值为1的点加入队列  for (int i = 0; i < m; i++) {  for (int j = 0; j < n; j++) {  if (matrix[i][j] == 1) {  queue.offer(new Point(i, j));  }  }  }  // 定义四个方向的偏移量  int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};  // 广度优先搜索  while (!queue.isEmpty()) {  Point current = queue.poll();  for (int[] dir : directions) {  int newX = current.x + dir[0];  int newY = current.y + dir[1];  // 检查新坐标是否有效且值为0  if (newX >= 0 && newX < m && newY >= 0 && newY < n && matrix[newX][newY] == 0) {  // 同化该点，并将其加入队列以便进一步同化其相邻的0  matrix[newX][newY] = 1;  queue.offer(new Point(newX, newY));  }  }  }  // 计算非1的元素个数  int count = 0;  for (int i = 0; i < m; i++) {  for (int j = 0; j < n; j++) {  if (matrix[i][j] != 1) {  count++;  }  }  }  return count;  }  public static void main(String[] args) {  Scanner scanner = new Scanner(System.in);  int m = scanner.nextInt();  int n = scanner.nextInt();  int[][] matrix = new int[m][n];  for (int i = 0; i < m; i++) {  for (int j = 0; j < n; j++) {  matrix[i][j] = scanner.nextInt();  }  }  // 确保起始点为1  if (matrix[0][0] != 1) {  matrix[0][0] = 1;  }  System.out.println(countNonOnes(matrix));  }  
}

4. 数组组成的最小数字(100)

public class Test {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);String[] str = scanner.nextLine().split(",");Arrays.sort(str, (a, b) -> {int lengthA = a.length();int lengthB = b.length();if (lengthA != lengthB) {return lengthA - lengthB;} else {return a.compareTo(b);}});StringBuilder stringBuilder = new StringBuilder();if (str.length == 1) {stringBuilder.append(str[0]);System.out.println(stringBuilder);} else if (str.length == 2) {String s1 = str[0] + str[1];String s2 = str[1] + str[0];stringBuilder.append(s1.compareTo(s2) < 0 ? s1 : s2);System.out.println(stringBuilder);} else {String[] res = Arrays.copyOf(str, 3);Arrays.sort(res);for (String r : res) {System.out.print(r);}}}
}

5. 垃圾信息拦截(200)

问题描述：按照如下规则进行垃圾信息识别，发送者A符合以下条件之一，则认为A为垃圾信息发送者：
A发送短信的接收者中，没有发过短信给A的人数L＞5
A发送的短信数-A接收的短信数M＞10
如果存在X，A发送给X的短信数-A接收到X的短信数N＞5

输入描述第一行为条目数目，接下来几行是具体的条目，每个条目是一对ID，第一个数字是发送者ID，后面的数字是接收者ID，中间空格隔开，所有的ID都是为无符号整型，ID最大值为100；
同一个条目中，两个ID不会相同（不会自己给自己发信息）
最后一行为指定ID

输出描述：输出该ID是否为垃圾短信发送者，并且按序列输出L M的值（由于N值不唯一，不需要输出）

例：输入：

15
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
1 11
1 12
1 13
1 14
14 1
1 15
1

输出：

true 13 13

输入：

15
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
1 11
1 12
1 13
1 14
14 1
1 15
2

输出：

false 0 -1

public class Test {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);List<Msg> msgList = new ArrayList<>();int count = Integer.parseInt(scanner.nextLine());for (int i=0;i<count;i++){String[] str = scanner.nextLine().split(" ");msgList.add(new Msg(Integer.parseInt(str[0]),Integer.parseInt(str[1])));}Boolean flag = false;int A = Integer.parseInt(scanner.nextLine());List<Msg> aSendList = msgList.stream().filter(msg -> msg.sendUser==A).collect(Collectors.toList());List<Integer> receiveFromA = aSendList.stream().map(msg -> {return msg.receiveUser;}).distinct().collect(Collectors.toList());List<Msg> otherSendList = msgList.stream().filter(msg -> msg.sendUser!=A).collect(Collectors.toList());List<Integer> aReceiveList = msgList.stream().filter(msg -> msg.receiveUser==A).map(msg -> {return msg.sendUser;}).collect(Collectors.toList());int M = aSendList.size()-aReceiveList.size();int L = aSendList.size();for (int i=0;i<otherSendList.size();i++){if (receiveFromA.contains(otherSendList.get(i).sendUser)){L-=1;}}if (M>10 || L>5){flag = true;System.out.print(flag + " " +   L + " " + M);}else {List<Msg> msgList1 = new ArrayList<>();for (int i = 0;i<receiveFromA.size();i++) {int Y = 0;int X = 0;int id = receiveFromA.get(i);Y = aReceiveList.stream().filter(m -> id ==m).collect(Collectors.toList()).size();X = aSendList.stream().filter(msg -> msg.receiveUser == id).collect(Collectors.toList()).size();if (Math.abs(X-Y)>5){flag = true;break;}}System.out.print(flag + " " +   L + " " + M);}}static  class  Msg{int sendUser;int receiveUser;Msg( int sendUser, int receiveUser){this.sendUser = sendUser;this.receiveUser = receiveUser;}}
}

6. 最长字符串的长度（二）（200）

package com.yue.test;import java.util.Scanner;/*** @Author: 夜雨* Create Date Time: 2024-09-02-8:26* Update Date Time:* @Version 1.0* @Description:* @see*/
public class Test {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);String s = scanner.next();int result = findLongestSubstringLength(s);System.out.println(result);}// 定义一个私有静态方法，用于找出满足条件的最长子字符串的长度private static int findLongestSubstringLength(String s) {int n = s.length();int maxLength = 0;for (int i = 0; i < n; i++) {int countL = 0, countO = 0, countX = 0;for (int j = 0; j < n; j++) {// 计算当前字符的索引，处理环形字符串的情况int index = (i + j) % n;// 获取当前字符char ch = s.charAt(index);if (ch == 'l') countL++;else if (ch == 'o') countO++;else if (ch == 'x') countX++;if (countL % 2 == 0 && countO % 2 == 0 && countX % 2 == 0) {maxLength = Math.max(maxLength, j + 1);}}}return maxLength;}
}