从Java中的阵列中删除重复项 [英] Removing duplicates from an Array in Java

问题描述

我有一个long类型的数组，我只是想编写一个可以查找和删除重复项的代码。它多少有些用，但是有一些错误。我不确定自己在做什么错。我非常感谢您的帮助。

I have an array of type long and I'm just trying to make a code that will find and remove duplicates. It somewhat works, but it has some bugs. I'm not sure what I'm doing wrong. I'd really appreciate the help.

我添加了数字：77、44、22、11、66、33、55、55、99、99、33， 0，0

I added the numbers : 77, 44, 22, 11, 66, 33, 55, 55, 99, 99, 33, 0, 0

，输出为：77、44、22、11、66、33、55、55、99、99

and the output is : 77, 44, 22, 11, 66, 33, 55, 55, 99, 99

因此它删除了33个重复项和两个0，并完全跳过了55和99。

so it erased the 33 duplicate and both 0's and completely skipped 55 and 99.

到目前为止，这是我的代码：

Here is my code so far:

nElems是数组的大小

nElems is the size of the array

public int noDups()
{
  int duplicates = 0;

    for(int i = 0; i<nElems; i++)
     {
        for(int j = i+1; j<nElems; j++)
         {
            if( i == j)
             {
                break;
             }
             else if (a[i] == a[j])
             {
                duplicates++;
                delete(j);
                nElems--;
             }
         }// end for j
      }// end for i

  return duplicates;

}// noDups()

我的删除看起来像这样：

My delete looks like this:

public boolean delete(long value)
{
  int j;

    for(j=0; j<nElems; j++) // look for it
     {
        if( value == a[j] )
            break;

        if(j==nElems) // can’t find it
          {
            return false;
          }
        else // found it
          {
             for(int k=j; k<nElems; k++) // move higher ones down
              {  
                  a[k] = a[k+1];
                  nElems--; // decrement size
                  return true;
              }
          }
     }// end for i
} // end delete()

推荐答案

public static class Node {
        int value;
        Node next;
        Node prev;

        public Node(int value)
        {
            this.value = value;
        }
    }

    public static class List {
        Node[] list = new Node[32];
        int size = 0;

        public void put(int value) {
            int index = value & 31;
            for (Node n = list[index]; n != null; n = n.next) {
                if (n.value == value) {
                    return;
                }
            }

            Node newNode = new Node(value);
            Node n = list[index];
            if (n != null) {
                n.prev = newNode;
                newNode.next = n;
            }
            list[index] = newNode;
            size++;
        }

        public void addAll(int[] array) {
            for (int x = 0; x < array.length; x++) {
                put(array[x]);
            }
        }

        public int[] toArray() {
            int[] array = new int[size];
            if (size != 0) {
                main:
                for (int b = 0, i = 0; b < list.length; b++) {
                    Node n = list[b];
                    for (; n != null; n = n.next) {
                        // Place this value in to our array.
                        array[i++] = n.value;
                        // We break because our index is larger than our
                        // available array size.
                        if (i >= size) {
                            break main;
                        }   
                    }
                }
            }
            return array;
        }
    }

    public static void main(String[] args) {
        List list = new List();
        int[] array = new int[] {77, 44, 22, 11, 66, 33, 55, 55, 99, 99, 33, 0, 0};
        list.addAll(array);
        System.out.println(Arrays.toString(list.toArray()));
    }

为您编写此代码。将很快完成您需要的一切！

Wrote this code out for you. Will do everything you need it todo and very fast!

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