背景 

针对于集合转换成map的时候出现value为空造成的空指针情况，进行解决

import lombok.Data;
import org.apache.commons.lang3.StringUtils;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Objects;
import java.util.Optional;
import java.util.stream.Collectors;

public class Main {
    public static void main(String[] args) {
        List<Student> list = new ArrayList<>();
        Student student = new Student();
        student.setName("AA");
        list.add(student);
        student = new Student();
        student.setName("BB");
        list.add(student);
        student = new Student();
        student.setName("BB");
        student.setPetName("B1");
        list.add(student);
        student = new Student();
        student.setName("CC");
        student.setPetName("C1");
        list.add(student);
        student = new Student();
        student.setName("CC");
        list.add(student);

        Map<String, String> collect = list.stream().collect(Collectors.toMap(x -> x.getName(), x -> x.getPetName(), (a, b) -> a));
        System.out.println(collect);

    }

    @Data
    static
    class Student {
        private String name;
        private String petName;
    }
}

在根据学生实体的名称作为key和昵称作为value的时候，当petName为空的时候会出现空指针 

Exception in thread "main" java.lang.NullPointerException
	at java.util.HashMap.merge(HashMap.java:1224)
	at java.util.stream.Collectors.lambda$toMap$58(Collectors.java:1320)
	at java.util.stream.ReduceOps$3ReducingSink.accept(ReduceOps.java:169)
	at java.util.ArrayList$ArrayListSpliterator.forEachRemaining(ArrayList.java:1374)
	at java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:481)
	at java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:471)
	at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:708)
	at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234)
	at java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:499)
	at com.aimsphm.console.symptom.service.impl.T.main(T.java:44)

解决方式

解决方式1

对空判断

Map<String, String> collect1 = list.stream().collect(Collectors.toMap(x -> x.getName(),
                x -> Objects.nonNull(x.getPetName()) ? x.getPetName() : "",
                (a, b) -> StringUtils.isNotBlank(a) ? a : b));
System.out.println(collect1);

 解决方式2

使用Optional对空值进行处理

Map<String, String> collect2 = list.stream().collect(Collectors.toMap(
                x -> Optional.ofNullable(x).map(Student::getName).orElse(""),
                x -> Optional.ofNullable(x).map(Student::getPetName).orElse(""),
                (a, b) -> StringUtils.isNotBlank(a) ? a : b));
System.out.println(collect2);

 解决方式3

使用 Collectors.reducing（）重新对分组之后的数据进行计算

Map<String, String> collect3 = list.stream().collect(Collectors.groupingBy(x ->
                x.getName(), HashMap::new, Collectors.reducing(null,
                x -> x.getPetName(), (a, b) -> Objects.nonNull(a) ? a : b)));
System.out.println(collect3);

 解决方式4

直接在collect方法中进行厘定

Map<String, String> collect4 = list.stream().collect(HashMap::new, (map, item) -> {
            if (!map.containsKey(item.getName()) || Objects.isNull(map.get(item.getName()))) {
                map.put(item.getName(), item.getPetName());
            }
        }, HashMap::putAll);
System.out.println(collect4);

 最终的结果输出为

## 输出的结果中有的使用了空字符串，可能需要根据实际需要进行调整
## 因为map是无序的，所以结果顺序并不一致

{CC=C1, BB=B1, AA=}
{CC=C1, BB=B1, AA=}
{CC=C1, BB=B1, AA=null}
{AA=null, BB=B1, CC=C1}

其中name没有考虑为空的情况， 可以使用filter进行过滤