这个问题是针对forEach的上下文提出的。
评论（在答案被接受后）：我接受了 @nullpointer 的答案，但它只在我的代码示例的情况下正确，而不适用于关于 reduce 可否“中断”的一般性问题。
问题：
但是，在reduce或collect中是否有一种方式可以“中断”操作，而不必遍历所有流元素？（这意味着需要在迭代时累积状态，因此我使用reduce或collect）。
简而言之：我需要遍历流的所有元素（元素为整数且按顺序从小到大），但要查看2个相邻元素并比较它们。如果它们之间的差大于1，则需要“中断”并停止“累积状态”，并且需要返回最后一个传递的元素。
抛出RuntimeException的变体和传递外部状态的变体-不适合我。
带有注释的代码示例：
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));
}
}