Question

这个类包含矩形列表，我需要找到面积最小的矩形。

我发现需要按区域比较矩形，但它具有双精度。我知道我的比较记得最后一个，但是我们如何在这里检查呢？

代码：

   /**
    * Gets the Rectangle with the smallest area
    * @return the rectangle with the smallest area or null if
    * there are no rectangles
    */
   public Rectangle smallestArea()
   {        
       if (list.size() == 0) return null;

       Rectangle smallest = list.get(0);           
       double smallestArea = smallest.getWidth() * smallest.getHeight();

       for (int i = 1; i < list.size(); i++) {
           Rectangle next = list.get(i);
           double nextArea = next.getWidth() * next.getHeight();

           if ((nextArea - smallestArea) < 0) smallest = next;             
       }

       return smallest;
   }

如何解决这个问题？

Answer 1

您还应该更新smallestArea局部变量（代码中还有一行）：

   public Rectangle smallestArea()
   {        
       if (list.size() == 0) return null;

       Rectangle smallest = list.get(0);           
       double smallestArea = smallest.getWidth() * smallest.getHeight();

       for (int i = 1; i < list.size(); i++) {
           Rectangle next = list.get(i);
           double nextArea = next.getWidth() * next.getHeight();

           if ((nextArea - smallestArea) < 0) { 
             smallest = next;         // <- Whenever you've updated smallest           
             smallestArea = nextArea; // <- Do not forget updating the smallestArea as well
           }  
       }

       return smallest;
   }

Answer 2

您的算法不起作用，因为您还需要更新smallestArea：

if ((nextArea - smallestArea) < 0) {
    smallest = next; 
    smallestArea = nextArea; // <<== Here
}

请注意，(nextArea - smallestArea) < 0是另一种说nextArea < smallestArea的方式，所以这看起来更干净：

if (nextArea < smallestArea) {
    smallest = next; 
    smallestArea = nextArea;
}

Answer 3

我愿意

double smallest = Double.POSITIVE_INFINITY;

// in the loop.
if (smallest > next)
    smallest = next;

Answer 4

除了记住最小的矩形之外，还需要记住它的大小。更改循环内的if：

if (nextArea < smallestArea) {
    smallest = next;
    smallestArea = nextArea
}

Answer 5

public Rectangle smallestArea()
{        
   if (list.size() == 0) return null;

   Rectangle smallest = list.get(0);           
   double smallestArea = smallest.getWidth() * smallest.getHeight();

   for (int i = 1; i < list.size(); i++) {
       Rectangle next = list.get(i);
       double nextArea = next.getWidth() * next.getHeight();

       if ((nextArea - smallestArea) < 0){
          smallest = next;             
          smallestArea = nextArea;
       }
   }

   return smallest;
}

Answer 6

这个怎么样？我想你可以尝试这种方式

    List<Rectangle> list=new ArrayList<>();
    List<Double> areaList=new ArrayList<>();
    for(Rectangle r:list){
        areaList.add(r.getHeight()*r.getHeight());
    }
    Collections.sort(areaList);
    System.out.println("Smallest "+areaList.get(0));

Answer 7

尝试，

int retval = Double.compare(nextArea, smallestArea);

if(retval < 0)
{
  System.out.println("smallestArea is greater than nextArea");
  smallest = next;
}

如何比较两个双数并找到更小的数

7 个答案: