Java的8个流，如何减少或收“破发”，没有抛出运行时异常？

Vladimir Nabokov 发表于 Java

121

纳博科夫：

这个问题已经被问的一个方面forEach。

评论（答案被接受后）：我接受@nullpointer的答案，但它是正确的只是在我的代码示例的情况下，而不是在大约减少的突破能力一般的问题..

问题：

但有一个方式reduce或collect以“破发”过早，没有通过所有的流元素回事？（这意味着我需要积累的状态，而迭代，所以我用reduce或collect）。

总之：我需要遍历流（元素是整数，从小下令大）的所有元素，但看看到2个相邻元素并加以比较，如果它们之间的差值小于1时，我需要“破发”和一站式“积累状态”，我需要返回最后传递的元素。

变异投掷RuntimeException和变种通过外部状态-对我不好。

代码示例与评论：

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));
}

}

也：

给定一个数组作为输入，在我看来，你要找的是这样的：

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
}

或者划痕，而将其转换为一个排序和筛选值列表如下：

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);
}

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