题目

除了最高和最低元素（值，而不是索引！）之外，对数组的所有数字求和（在F＃和Haskell中得到一个列表）。
（最高/最低元素在每个边缘分别只有一个元素，即使有多个元素具有相同的值！）

示例：

{ 6, 2, 1, 8, 10 } => 16
{ 1, 1, 11, 2, 3 } => 6

如果array为空，null或None，或者如果只存在1个Element，则返回0.
注意：在C ++中，使用空向量。在C中没有null。;-)

-- There's no null in Haskell, therefore Maybe [Int] is used. Nothing represents null.

测试用例：

import org.junit.Test;
import static org.junit.Assert.assertEquals;
import org.junit.runners.JUnit4;

public class SolutionTest {
    @Test
    public void BasicTests() {
      assertEquals(16, Kata.sum(new int[] { 6, 2, 1, 8, 10}));     
    }
    @Test
    public void BasicTests1() {
      assertEquals(0, Kata.sum(new int[] {}));     
    }
}

解题

My

public class Kata
{
  public static int sum(int[] numbers)
  {
      if(numbers == null){
        return 0;
      }
      for (int i = 0; i < numbers.length - 1; i++) {
          for (int j = 0; j < numbers.length - 1 - i; j++) {
              if (numbers[j] > numbers[j + 1]) {
                  int tmp = numbers[j];
                  numbers[j] = numbers[j + 1];
                  numbers[j + 1] = tmp;
              }
          }
      }
      int sum = 0;
      for (int j = 1; j < numbers.length - 1; j++) {
          sum += numbers[j];
      }
      return sum;
  }
}

Other

public class Kata
{
  public static int sum(int[] numbers)
  {
    if (numbers == null || numbers.length == 0 || numbers.length == 1) return 0;
    int min,max,sum;
    sum = min = max = numbers[0];
    for (int i = 1; i < numbers.length; i++)
    {
      sum += numbers[i];
      if (numbers[i] < min) min = numbers[i];
      if (numbers[i] > max) max = numbers[i];
    }
    return sum - min - max;
  }
}

import static java.util.stream.IntStream.of;

public class Kata {

  public static int sum(int[] numbers) {
    return (numbers == null || numbers.length <= 2) ? 0 : of(numbers).sum() - of(numbers).max().getAsInt() - of(numbers).min().getAsInt();
  }
}

import java.util.Arrays;

public class Kata
{
  public static int sum(int[] numbers)
  {
    if(numbers == null || numbers.length < 2) return 0;
    Arrays.sort(numbers);
    return Arrays.stream(numbers).skip(1).limit(numbers.length-2).sum();
  }
}

后记

Other的第一个确实聪明呀。