描述

Midterm 2 Review Solutions Streams Intermediate Operations to Know 1. stream(): Converts a collection into a stream 2. sorted(Comparator c): Sorts the stream using c, which can also be a lambda 3. map(Function f)(): Applies f to each element in the stream 4. filter(Predicate p): Tests each element with p to see if it should remain in the stream Terminal Operations to Know 1. reduce(BinaryOperator accumulator): Applies the accumulator and combines the elements in the stream, two at a time. Returns an Optional. 2. collect(Collector c): Turns the stream into a collection. To a List: Collectors.toList() To a Set: Collectors.toSet() To a Map: Collectors.grooupingBy(Function f) 3. forEach(Consumer action) Practice Problems Given the following classes, write methods to perform the desired operations. You may only use one semicolon per method. public class Student { public int getLabSection() { ... } public int getAge() {...} public Group getGroup {...} } public class Group { public int size() {...} public Group merge(Group other) { ... } } 1 1. Collect all the unique lab sections from a list of students. There should not be duplicates. public Collection<Integer> uniqueSections(List<Student> lst) { return lst.stream() .map(Student::getLabSection) .collect(Collectors.toSet()); } 2. Each TA submits a list of their students. Write a method that will map each lab section to the students in that section whose group is larger than size 4. Assume there is always at least one student in each section whose group size is larger than 4. public Map<Integer, List<Student>> grpMap(List<List<Student>> lst) { return lst.stream() .map(o -> o.stream() .filter(getGroup.size() > 4) .collect(Collectors.toList())) .collect(Collectors.groupingBy( o -> o.get(0).getLabSection())); } 3. Each TA submits a list of their students. Write a method that will get the number of students who are over age 20. public int over20 (List<List<Student>> lst) { return lst.stream() .map(o-> o.stream() .f

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