如何拆分奇数和偶数和和都在一个集合使用流方法的Java
public class SplitAndSumOddEven {
public static void main(String[] args) {
// Read the input
try (Scanner scanner = new Scanner(System.in)) {
// Read the number of inputs needs to read.
int length = scanner.nextInt();
// Fillup the list of inputs
List<Integer> inputList = new ArrayList<>();
for (int i = 0; i < length; i++) {
inputList.add(scanner.nextInt());
}
// TODO:: operate on inputs and produce output as output map
Map<Boolean, Integer> oddAndEvenSums = inputList.stream(); // Here I want to split odd & even from that array and sum of both
// Do not modify below code. Print output from list
System.out.println(oddAndEvenSums);
}
}
}
您可以使用Collectors.分隔符来做您想要的：
Map<Boolean, Integer> result = inputList.stream().collect(
Collectors.partitioningBy(x -> x%2 == 0, Collectors.summingInt(Integer::intValue)));
生成的映射包含true键中的偶数之和和false键中的奇数之和。
在两个单独的流操作中执行它是最简单（也是最干净的）的，如下所示：
public class OddEvenSum {
public static void main(String[] args) {
List<Integer> lst = ...; // Get a list however you want, for example via scanner as you are.
// To test, you can use Arrays.asList(1,2,3,4,5)
Predicate<Integer> evenFunc = (a) -> a%2 == 0;
Predicate<Integer> oddFunc = evenFunc.negate();
int evenSum = lst.stream().filter(evenFunc).mapToInt((a) -> a).sum();
int oddSum = lst.stream().filter(oddFunc).mapToInt((a) -> a).sum();
Map<String, Integer> oddsAndEvenSumMap = new HashMap<>();
oddsAndEvenSumMap.put("EVEN", evenSum);
oddsAndEvenSumMap.put("ODD", oddSum);
System.out.println(oddsAndEvenSumMap);
}
}
我所做的一个改变是使生成的Map成为Map
List<Integer> numberList = Arrays.asList(1,3,4,60,12,22,333456,1239);
Map<Boolean, List<Integer>> evenOddNumbers = numberList.stream()
.collect(Collectors.partitioningBy( x -> x%2 == 0));
System.out.println(evenOddNumbers);
输出：
{false=[1,3,1239]， true=[4,60,12,22,333456]}
如果想把它分开，使用下面的一个：
List<Integer> evenNums = partitions.get(true);
List<Integer> oddNums = partitions.get(false);
