Question

对于我的java作业，我正在努力编写一个递归合并排序类。截至目前，我有3个方法，一个“驱动”方法来启动递归，递归mergeSort方法和merge方法。根据我更改的变量，我的输出是一个全零的数组或我的原始数组以相同的顺序。唯一的原因是原始mergeSort方法必须采用一个数组，merge方法不能返回任何内容。任何帮助都非常感激

import java.util.Arrays;
public class merge2 {
    public static void main(String[] args){
        int []a={22,45,1,4,89,7,0};
        mergeSort(a);
        System.out.println(Arrays.toString(a));                 
    }

    public static void mergeSort(int [] a){
        mergeSort(a,0,a.length-1);
    }

    public static void mergeSort(int []a, int beg,int end){
        if(beg<end){
            int mid=(beg+end)/2;
            mergeSort(a,beg,mid);
            mergeSort(a,mid+1,end);
            merge(a,beg,mid,end);
        }
    }

    private static void merge(int []a, int beg, int middle, int end){
        int [] d=new int[a.length];
        int mid=middle+1; //start of second half of array
        for(int i=0;i<d.length;i++){
            if(beg<=middle && mid<=end){  
                if(a[beg]<=a[mid]) {
                d[i]=a[beg];
                beg++;
                } else if(a[mid]<=a[beg]){
                        d[i]=a[mid];
                        mid++;
                }
            }else if(beg>middle){ 
                d[i]=a[mid];
                mid++;
            }else if(mid==a.length){
                d[i]=a[beg];
                beg++;
            }
        }
        for(int w=0;w<d.length;w++){
            a[w]=d[w];
        }
    }
}

Answer 1

这是mergeSort方法的伪代码。我看到你忽略了一两个元素的基本情况：

if (sublist has only one value)
   do nothing
else if (sublist has two values)
   switch if necessary
else    // recursion, divide list into two halves
   Find midpoint of current sublist
   Call mergeSort and process left sublist
   Call mergeSort and process right sublist
   merge left and right sublists

对于您的合并方法，它正在创建一个新数组，而不是修改现有数组。我建议使用ArrayLists来实现它：

private void merge(int[] a, int first, int mid, int last)
  {
    ArrayList<Integer> left = new ArrayList<Integer>(mid - first + 1), right = new ArrayList<Integer>(last - mid);
    for(int m = first; m <= mid; ++m)   //initialize left sublist
    {
      left.add(a[m]);
    }
    for(int m = mid + 1; m <= last; ++m)    //initialize right sublist
    {
      right.add(a[m]);
    }
    int i = first;
    while(!left.isEmpty() || !right.isEmpty())  //while either list has an element
    {
      if(left.isEmpty())
      {
        a[i++] = right.remove(0);
      }
      else if(right.isEmpty())
      {
        a[i++] = left.remove(0);
      }
      else if (left.get(0) < right.get(0))
      {
        a[i++] = left.remove(0);
      }
      else
      {
        a[i++] = right.remove(0);
      }
    }
  }

Answer 2

好的我修复了你的解决方案。您的主要问题是在标有//<= here的行上。当mid在结束索引上运行时，您没有将a的值分配给d，因此它会被0填充。您必须将==替换为>=才能解决此问题我还用索引修复了你的工作。您不必在每个级别上运行整个数组。你的复杂性也会受到这种影响。我认为它是关于O(n^2)。仅运行在此递归级别上处理的部分数组就足以保持O(nlog(n))复杂性。

固定算法如下

public static void main(String[] args){
    int []a={22,45,1,4,89,7,0};
    mergeSort(a);
    System.out.println(Arrays.toString(a));

}

public static void mergeSort(int [] a){
    mergeSort(a,0,a.length-1);
}

public static void mergeSort(int []a, int beg,int end){
    if(beg<end){
        int mid=(beg+end)/2;
        mergeSort(a,beg,mid);
        mergeSort(a,mid+1,end);
        merge(a,beg,mid,end);
    }
}

private static void merge(int []a, int beg, int middle, int end){
    int [] d=new int[a.length];
    int mid=middle+1; //start of second half of array
    int beg1=beg;
    for(int i=beg1;i<=end;i++){
        if(beg<=middle && mid<=end){  
            if(a[beg]<=a[mid]) {
            d[i]=a[beg];
            beg++;
            } else if(a[mid]<=a[beg]){
                    d[i]=a[mid];
                    mid++;
            }
        }else if(beg>middle){         
            d[i]=a[mid];
            mid++;
        }else if(mid>=end){ //<= here
            d[i]=a[beg];
            beg++;
        }
    }
    for(int w=beg1;w<=end;w++){
        a[w]=d[w];
    }
}

合并排序有问题

2 个答案: