求和java arraylist的所有元素

Question

如果我有:ArrayList<Double> m = new ArrayList<Double>(); 使用里面的double值,我该如何添加所有ArrayList元素？

public double incassoMargherita()
{
 double sum = 0;
 for(int i = 0; i < m.size(); i++)
 {          
 }
 return sum;
}

如？

Answer 1

两种方式:

使用索引:

double sum = 0;
for(int i = 0; i < m.size(); i++)
    sum += m.get(i);
return sum;

使用"for each"样式:

double sum = 0;
for(Double d : m)
    sum += d;
return sum;

Answer 2

Java的8+版本Integer，Long，Double和Float

    List<Integer> ints = Arrays.asList(1, 2, 3, 4, 5);
    List<Long> longs = Arrays.asList(1L, 2L, 3L, 4L, 5L);
    List<Double> doubles = Arrays.asList(1.2d, 2.3d, 3.0d, 4.0d, 5.0d);
    List<Float> floats = Arrays.asList(1.3f, 2.2f, 3.0f, 4.0f, 5.0f);

    long intSum = ints.stream()
            .mapToLong(Integer::longValue)
            .sum();

    long longSum = longs.stream()
            .mapToLong(Long::longValue)
            .sum();

    double doublesSum = doubles.stream()
            .mapToDouble(Double::doubleValue)
            .sum();

    double floatsSum = floats.stream()
            .mapToDouble(Float::doubleValue)
            .sum();

    System.out.println(String.format(
            "Integers: %s, Longs: %s, Doubles: %s, Floats: %s",
            intSum, longSum, doublesSum, floatsSum));

15、15、15.5、15.5

Answer 3

我没有测试过,但它应该工作.

public double incassoMargherita()
{
    double sum = 0;
    for(int i = 0; i < m.size(); i++)
    {
        sum = sum + m.get(i);
    }
    return sum;
}

Answer 4

使用Java 8 流:

double sum = m.stream()
    .mapToDouble(a -> a)
    .sum();

System.out.println(sum); 

Answer 5

并不是很难，只需使用m.get(i)从列表中获取值即可。

public double incassoMargherita()
{
    double sum = 0;
    for(int i = 0; i < m.size(); i++)
    {
        sum += m.get(i);
    }
    return sum;
}