在数组中查找重复项并仅打印一次 [英] Finding Duplicates in Array and printing them only Once

问题描述

我正在尝试遍历我的数组并找到所有重复多次的数字：

I am trying to loop through my array and find all the numbers that are repeating more than once:

EG：如果有 1 1 2 3 4

应打印说1次重复多次

这是我的代码，到目前为止我尝试了，但它打印所有重复并继续，如果有 4 4 4 4 3 6 5 6 9 ，它将打印所有的4，但我不想要：

Here is my code and so far what I have tried, however it prints all duplicates and keep going, if there is 4 4 4 4 3 6 5 6 9, it will print all the 4's but i dont want that:

class average {

 public static void main(String[] args) throws IOException {

    int numOfLines = 0;
    int sum = 0, mean = 0, median = 0, lq = 0, uq = 0;
    int[] buffer;

    File myFile = new File("num.txt");
    Scanner Scan = new Scanner(myFile);

    while(Scan.hasNextLine()) {
        Scan.nextLine();
        numOfLines++;
    }
    Scan.close();
    Scan = new Scanner(myFile);

    System.out.println("Number Of Lines: " + numOfLines);

    buffer = new int[numOfLines];

    for(int i=0; i<numOfLines; i++) {
        buffer[i] = Scan.nextInt();
    }
    Scan.close();
    Scan = new Scanner(myFile);

    for(int i=0; i<buffer.length; i++) {
        sum = sum+i;
        mean = sum/numOfLines;
    }
    System.out.println("Sum: " + sum);
    System.out.println("Mean: " + mean);

    for(int i=0; i<buffer.length; i++) {
        for(int k=i+1; k<buffer.length; k++) {
            if(buffer[k] == buffer[i]) {
                System.out.println(buffer[k]);
            }
        }
    }

推荐答案

只需将您发现重复的数字添加到某些结构中，例如 HashSet 或 HashMap ，这样您就可以找到它稍后你会发现另一个重复。

Just add the number you will find duplicated to some structure like HashSet or HashMap so you can find it later when you will detect another duplication.

Set<Integer> printed = new HashSet<Integer>();

  for(int i=0; i<buffer.length; i++) {
    for(int k=i+1; k<buffer.length; k++) {
      if(buffer[k] == buffer[i]) {
        Integer intObj = new Integer(buffer[k]);
        if (!printed.contains(intObj)) {
          System.out.println(buffer[k]);
          printed.add(intObj);
        }
        break;
       }
    }
  }

更好的O（ n）alghorithm：

Set<Integer> printed = new HashSet<Integer>();

  for(int i=0; i<buffer.length; i++) {
    if (!printed.add(new Integer(buffer[i])) {
       System.out.println(buffer[i]);
    }
  }

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