Java 8流，如何在reduce或collect中“中断”而不抛出运行时异常？

Question

这个问题是在 的背景下提出的forEach。

评论（答案被接受后）：我接受了@nullpointer的答案，但它仅在我的代码示例的上下文中才是正确的，而不是在关于reduce的可破坏性的一般问题中。

问题：

但是有没有一种方法可以reduce在collect不遍历所有流元素的情况下提前“中断”？（这意味着我需要在迭代时累积状态，所以我使用reduce或collect）。

简而言之：我需要迭代流的所有元素（元素是整数并从小到大排序），但是查看 2 个相邻元素并比较它们，如果它们之间的差异大于 1，我需要“中断”并停止“累积状态”，我需要返回最后传递的元素。

抛出 a 的变体RuntimeException和传递外部状态的变体 - 对我来说不好。

带注释的代码示例：

public class Solution {

public int solution(int[] A) {

    Supplier<int[]> supplier = new Supplier<int[]>() {
        @Override
        public int[] get() {
            //the array describes the accumulated state:
            //first element in the array , if set > 0, means  - the result is achieved, we can stop iterate over the rest elements
            //second element in the array will represent the "previous element" while iterating the stream
            return new int[]{0, 0};
        }
    };

    //the array in accumulator describes the accumulated state:
    //first element in the array , if set > 0, means  - the result is achieved, we can stop iterate over the rest elements
    //second element in the array will represent the "previous element" while iterating the stream
    ObjIntConsumer<int[]> accumulator = new ObjIntConsumer<int[]>() {
        @Override
        public void accept(int[] sett, int value) {
            if (sett[0] > 0) {
                ;//do nothing, result is set
            } else {
                if (sett[1] > 0) {//previous element exists
                    if (sett[1] + 1 < value) {
                        sett[0] = sett[1] + 1;
                    } else {
                        sett[1] = value;
                    }
                } else {
                    sett[1] = value;
                }
            }
        }
    };

    BiConsumer<int[], int[]> combiner = new BiConsumer<int[], int[]>() {
        @Override
        public void accept(int[] sett1, int[] sett2) {
            System.out.println("Combiner is not used, we are in sequence");
        }
    };

    int result[] = Arrays.stream(A).sorted().filter(value -> value > 0).collect(supplier, accumulator, combiner);
    return result[0];
}


/**
 * We have an input array
 * We need order it, filter out all elements that <=0 (to have only positive)
 * We need find a first minimal integer that does not exist in the array
 * In this example it is 5
 * Because 4,6,16,32,67 positive integers array is having 5 like a minimum that not in the array (between 4 and 6)
 *
 * @param args
 */
public static void main(String[] args) {
    int[] a = new int[]{-2, 4, 6, 16, -7, 0, 0, 0, 32, 67};
    Solution s = new Solution();
    System.out.println("The value is " + s.solution(a));
}

}

Answer 1

给定一个数组作为输入，在我看来你正在寻找这样的东西：

int stateStream(int[] arr) {
    return IntStream.range(0, arr.length - 1)
            .filter(i -> arr[i + 1] - arr[i] > 1) // your condition
            .mapToObj(i -> arr[i])
            .findFirst() // first such occurrence
            .map(i -> i + 1) // to add 1 to the point where the cehck actually failed
            .orElse(0); // some default value
}

或者从头开始，同时将其转换为排序和过滤的值列表，如下所示：

int stateStream(int[] arr) {
    List<Integer> list = Arrays.stream(arr)
            .boxed().sorted()
            .filter(value -> value > 0)
            .collect(Collectors.toList());
    return IntStream.range(0, list.size() - 1)
            .filter(i -> list.get(i + 1) - list.get(i) > 1)
            .mapToObj(list::get)
            .findFirst()
            .map(i -> i + 1)
            .orElse(0);
}