Java - 过滤掉最低整数的汇总列表

Question

所以，我有一个 int[] arr 并想在过滤掉最小 int 的同时使用流对其求和。我有一个使用两个流的解决方案，它有效但似乎无效。有没有办法只用一个流来做到这一点？

代码:

int min = Arrays.stream(arr)
                    .min()
                    .getAsInt();
int sum = Arrays.stream(arr)
                       .filter(i -> i != min)
                       .sum();

Answer 1

以下代码使用 IntSummaryStatistics应该可以解决问题。

public static void main(String[] args) {
    int[] arr = {1, 2, 3, 4, 5, 6};
    IntSummaryStatistics stats = Arrays.stream(arr).summaryStatistics();
    int sum = (int) stats.getSum() - stats.getMin();
}

来自 IntSummaryStatistics 的文档:

A state object for collecting statistics such as count, min, max, sum, and average.

...

This computes, in a single pass, the count of people, as well as the minimum, maximum, sum, and average of their number of dependents.

编辑:如果您想删除所有具有最小值的元素:

    int[] arr = {1, 2, 3, 1, 1, 1};

    TreeMap<Integer, Integer> map = Arrays.stream(arr).boxed()
        .collect(toMap(
            v -> v,
            v -> 1,
            Integer::sum,
            TreeMap::new
        ));

    map.remove(map.firstKey());
    int sum = map.entrySet().stream().mapToInt(e -> e.getKey() * e.getValue()).sum();
    System.out.println(sum);

或

    List<Integer> list = Arrays.stream(arr).sorted().boxed().collect(toList());
    Integer min = list.get(0);
    int sum2 = list.stream().mapToInt(i -> i).dropWhile(min::equals).sum();
    System.out.println(sum2);