JAVA CERTIFICATION QUESTION: Search stream data using the findFirst, findAny, and anyMatch methods
Streams are complicated. With the right approach, they can be very efficient too.
Streams are complicated. With the right approach, they can be very efficient too.
Given the following code:
List src = List.of("Java 11", "Exam");
Which code fragment determines if the word “Java” is present in the src list in the most computationally efficient way? Choose one.
A.
List res = List.of();
src.stream()
.peek(v -> { if (v.contains("Java")) res.add(v); })
.count();
var a = (res.size() > 0);
B.
var a = src.stream().filter(v -> v.contains("Java")).findAny();
C.
var a = src.stream().anyMatch(v -> v.contains("Java"));
D.
var a = src.stream()
.filter(v -> v.contains("Java"))
.collect(Collectors.toList());
The answer is C.