Question

I m 试图建立一个用扼杀而不是ger器的宾平搜索方案。问题在于,我不知道如何使数量少而少,这样就显得微不足道。

例如

阵列的阵列少于扼杀值。

/**
   The StringBinarySearcher class provides a public static
   method for performing a binary search on an String array.
*/



public class StringBinarySearcher
{
   /**
      The search method performs a binary search on an String
      array. The array is searched for the number passed to
      value. If the number is found, its array subscript is
      returned. Otherwise, -1 is returned indicating the
      value was not found in the array.
      @param numbers The array to search.
      @param value The value to search for.
   */



   public static int search(String[] numbers, String value)
   {
      int first;       // First array element
      int last;        // Last array element
      int middle;      // Mid point of search
      int position;    // Position of search value
      boolean found;   // Flag

      // Set the inital values.
      first = 0;
      last = numbers.length - 1;
      position = -1;
      found = false;

      // Search for the value.
      while (!found && first <= last)
      {
         // Calculate mid point
         middle = (first + last) / 2;

         // If value is found at midpoint...
         if (numbers[middle] == value)
         {
            found = true;
            position = middle;
         }

         // else if value is in lower half...
         // needs array to be less then the string value?, without using equality regulators
         else if (numbers[middle].compareTo(numbers[middle +1]) > 0)
            last = middle - 1;
         // else if value is in upper half....
         else
            first = middle + 1;
      }

      // Return the position of the item, or -1
      // if it was not found.
      return position;
   }
}

Answer 1

第一项比较:

if (numbers[middle] == value)

<>not > 使用=营运人。页: 1 您应使用: 方法或 compareTo 。

下面的比较是:

// else if value is in lower half...
// needs array to be less then the string value?, without using equality regulators
else if (numbers[middle].compareTo(numbers[middle +1]) > 0)

根据你的评论,你重新检查,看看<代码>> 数值/代码”在阵列的下半部分,但你的代码将阵列要素与下一个阵列要素相比较。为了与评论意见保持一致,应当:

else if (value.compareTo(numbers[middle]) < 0)

另外,请指出,在比较代表你重新行事方式的“强项”时,你会取得一些奇怪的结果。。 • 地理学。例如,这意味着“5”的评价大于“11”。

Answer 2

你的问题是比较操作员(=)。比较操作者仅对 Java的原始数据类型作了明确界定。浏览是一种类别(不是原始数据类型)。因此,您将需要使用“Sting s>codequalitys(String)方法加以比较。

If you want to compare them as numbers, then you will need to parse them into Integers. For this, you can use Integer.parseInt(String) and then compare the Integers.

Answer 3

对于原始数据类型,这种工程是罚款的。不是用于 Str。

= 用于检查两个物体的参照点是否相同。

String strName1 = "Me";
String strName2 = new String("Me");

strName1 == strName2 is false. Since they are referring to two different objects.

你们可以采用平等的方法进行比较。

if (strName2 .equals(strName2 )) {
    System.out.println("Me and Me are same :P");
}

友情链接