我有这个代码
List<Integer> numbers = new ArrayList<>(30, 25, 17, 12 ,8, 5, 3, 2));
List<Integer> indices = new ArrayList<>(5, 3, 2));
Integer newNumber = 1;
for (int i = indices.size() - 1; i >= 0; i--) {
newNumber *= (numbers.get(indices.get(i)));
}
newNumber
将是:5*12*17 = 1020
可以使用stream reduce吗?
稍后我需要从原始数字中删除索引(我在过滤器中考虑,但目标是索引而不是 Integer 对象)。
List<Integer> newNumbers = new ArrayList<>(numbers);
for (int i = indices.size() - 1; i >= 0; i--) {
newNumbers.remove(indices.get(i).intValue()); // Remove position
}
或者,我在考虑这段代码。
List<Integer> newNumbers2 = new ArrayList<>();
for (int i = 0; i < numbers.size(); i++) {
if (!indices.contains(i)) {
newNumbers2.add(numbers.get(i));
}
}
是否可以使用流来实现?
谢谢。
最佳答案
是的,你可以使用简单的减少来做到这一点。在这种情况下,归约的标识元素为 1。
int product = indices.stream().mapToInt(numbers::get).reduce(1, (n1, n2) -> n1 * n2);
后一个问题的答案是,
Set<Integer> indicesSet = new HashSet<>(indices);
List<Integer> newNumbers = IntStream.range(0, numbers.size())
.filter(n -> !indicesSet.contains(n))
.mapToObj(n -> numbers.get(n))
.collect(Collectors.toList());
In mathematics, an identity element is a special type of element of a set with respect to a binary operation on that set, which leaves any element of the set unchanged when combined with it.
关于Java 8,流与另一个基于列表索引,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56014539/