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  • 问题:从data.bat文件中排序,找到数量最大的10个数字,以println的方式分别输出。

    问题:从data.bat文件中排序,找到数量最大的10个数字,以println的方式分别输出。

    要求:
    0.时间尽量短
    1.类名为Test.java
    2.无包名
    3.无第三方引用
    4.可直接javac编译和java运行
    5.java命令行运行class,参数为data.bat文件路径,例如,java Test ./data.bat
    6.jdk1.8
    7.文件字符集,utf-8
    8.不能有异常和警告

    data.bat  地址:https://github.com/TabMM/drafts/tree/master/data.bat

    废话不多说 直接上代码 (未经允许,不得转载)

    方法一(本人写):

    public class Test {
    public static void main(String[] args) {
    long start = System.currentTimeMillis();
    readFileByLines();
    System.err.print("使用时间" + (System.currentTimeMillis() - start) + "ms");
    }

    /**
    * 以行为单位读取文件,常用于读面向行的格式化文件
    */

    public static void readFileByLines() {
    int[] s = new int[]{0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
    File file = new File("F:\IdeaProjects\snowflake\src\main\java\com\example\snowflake\data.dat");
    BufferedReader reader = null;
    try {
    reader = new BufferedReader(new FileReader(file));
    String tempString = null;
    int line = 1;
    // 一次读入一行,直到读入null为文件结束
    while ((tempString = reader.readLine()) != null) {
    int temp = Integer.parseInt(tempString);
    int t = temp > s[0] ? jh(0, temp, s) : (temp > s[1] ? jh(1, temp, s) : (temp > s[2] ? jh(2, temp, s) : (temp > s[3] ? jh(3, temp, s) : (temp > s[4] ? jh(4, temp, s) : (temp > s[5] ? jh(5, temp, s) : (temp > s[6] ? jh(6, temp, s) : (temp > s[7] ? jh(7, temp, s) : (temp > s[8] ? jh(8, temp, s) : (temp > s[9] ? jh(9, temp, s) : 0)))))))));
    line++;
    }
    System.out.println(s[0] + " " + s[1] + " " + s[2] + " " + s[3] + " " + s[4] + " " + s[5] + " " + s[6] + " " + s[7] + " " + s[8] + " " + s[9]);
    reader.close();
    } catch (IOException e) {
    e.printStackTrace();
    } finally {
    if (reader != null) {
    try {
    reader.close();
    } catch (IOException e1) {
    }
    }
    }
    }

    private static int jh(int index, int temp, int[] s) {
    int a = index;
    for (int i = 0; i < 9 - index; i++) {
    stamp(i, s);
    }
    s[a] = temp;
    return 0;
    }

    private static void stamp(int index, int[] s) {
    s[9 - index] = s[8 - index];
    }

    }

    方法二(@小和写):
    public class Demo {
    public static void main(String[] args) throws Exception {
    long start = System.currentTimeMillis();
    TreeMap<Integer,Integer> treeMap = new TreeMap<>(new Comparator<Integer>() {
    @Override
    public int compare(Integer o1, Integer o2) {
    return o2.compareTo(o1);
    }
    });
    BufferedReader br=new BufferedReader(new FileReader("F:\IdeaProjects\snowflake\src\main\java\com\example\snowflake\data.dat"));//读取文件
    //输出
    for(String s=br.readLine();s!=null;s=br.readLine()) {
    treeMap.put(Integer.parseInt(s), Integer.parseInt(s));
    if(treeMap.size() > 10){
    treeMap.remove(treeMap.lastKey());
    }
    }
    Iterator<Map.Entry<Integer,Integer>> it = treeMap.entrySet().iterator();
    while(it.hasNext()) {
    Map.Entry<Integer,Integer> entry = it.next();
    System.out.println(entry.getKey() + " : " + entry.getValue());
    }
    System.err.print( "使用时间"+(System.currentTimeMillis() - start) +"ms");
    }
    }
    方法三(@咿呀咿呀哟写):
    public class Test {

    public static int N = 10; // Top10
    public static int arr[] = new int[N];
    //数组长度
    public static int len = arr.length;
    //堆中元素的有效元素 heapSize<=len
    public static int heapSize = len;

    public static void main(String[] args){
    long start = System.currentTimeMillis();
    int _150M = 1024*1024*150;
    File fin = new File("F:\IdeaProjects\snowflake\src\main\java\com\example\snowflake\data.dat");
    FileChannel fcin = null;
    try {
    fcin = new RandomAccessFile(fin, "r").getChannel();
    } catch (FileNotFoundException e) {
    System.out.println("系统找不到指定文件");
    return;
    }
    ByteBuffer rBuffer = ByteBuffer.allocate(_150M);
    readFileByLine(_150M, fcin, rBuffer);
    System.err.print( "使用时间"+(System.currentTimeMillis() - start) +"ms");
    }
    public static void readFileByLine(int bufSize, FileChannel fcin,
    ByteBuffer rBuffer) {
    String enterStr = " ";
    try {
    byte[] bs = new byte[bufSize];
    String tempString = null;
    while (fcin.read(rBuffer) != -1) {
    int rSize = rBuffer.position();
    rBuffer.rewind();
    rBuffer.get(bs);
    rBuffer.clear();
    tempString = new String(bs, 0, rSize);
    int fromIndex = 0;
    int endIndex = 0;
    int i = 0 ;
    while ((endIndex = tempString.indexOf(enterStr, fromIndex)) != -1) {
    String line = tempString.substring(fromIndex, endIndex);
    fromIndex = endIndex + 1;
    if (i<10){
    arr[i]= Integer.valueOf(line);
    i++;
    continue;
    }
    //构建小顶堆
    buildMinHeap();
    if(Integer.valueOf(line) > arr[0]){
    arr[0] = Integer.valueOf(line);
    minHeap(0);
    }
    }
    print();
    }
    } catch (IOException e) {
    System.out.println("读取文件失败");
    return;
    }
    }
    /**
    * 自底向上构建小堆
    */
    public static void buildMinHeap(){
    int size = len / 2;
    for(int i = size;i>=0;i--){
    minHeap(i);
    }
    }

    /**
    * i节点为根及子树是一个小堆
    * @param i
    */
    public static void minHeap(int i){
    int l = left(i);
    int r = right(i);
    int index = i;
    if(l<heapSize && arr[l]<arr[index]){
    index = l;
    }
    if(r<heapSize && arr[r]<arr[index]){
    index = r;
    }
    if(index != i){
    int t = arr[index];
    arr[index] = arr[i];
    arr[i] = t;
    //递归向下构建堆
    minHeap(index);
    }
    }

    /**
    * 返回i节点的左孩子
    * @param i
    * @return
    */
    public static int left(int i){
    return 2*i;
    }

    /**
    * 返回i节点的右孩子
    * @param i
    * @return
    */
    public static int right(int i){
    return 2*i+1;
    }
    /**
    * 打印
    */
    public static void print(){
    for (int i = arr.length -1; i >=0 ; i--) {
    System.out.println(arr[i]);
    }
    }
    }
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  • 原文地址:https://www.cnblogs.com/ludengfu/p/9475901.html
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