List整理 - 技术栈

List

一个list转为二维数组，五个一组

List<List<String>> lists = Lists.partition(list,5);
删除list中的元素删除下标以及定位到遍历的位置

for(int i = 0, len = list.size(); i < len; i++){
if(list.get(i) == 1){
list.remove(i);
len--;
i--;
}
}

List转数组

// 要转换的list集合
List<String> testList = new ArrayList<String>(){{add("aa");add("bb");add("cc")}}

// 使用toArray(T[] a)方法
String[] array2 = testList.toArray(new Strinng[testList.size()]);

List对象类型转换

List<Response1> collect = response2List.stream().map(item -> ConvertHelper.convert(item, Response1.class)).collect(Collectors.toList());
List和List相互转换

// string转long
List<Long> longList = stringList.stream.map(Long::valueOf).collect(Collectors.toList());
// long转string
List<String> stringList = longList.stream.map(String::valueOf).collect(Collectors.toList());
list取其中一个字段组成List

List<Long> numberList = list.stream().map(Response::getOrgId).collect(Collectors.toList());
List取差集

allDataList.removeAll(removeDataList);
多个List取交集

List<Integer> A = new ArrayList<>(Arrays.asList(1,2,5,8));
List<Integer> B = new ArrayList<>(Arrays.asList(5,7,8,9));
List<Integer> C = new ArrayList<>(Arrays.asList(5,8,9));
List<List<Integer>> D = new ArrayList<>();
D.add(A);
D.add(B);
D.add(C);
List<Integer> E = D.get(0);
D.forEach(integers ->{
E.retailAll(integers);
});

// E就是最后的交集结果
在list后追加list / 在list中添加list

list1.addAll(list2);
list的两种分组方式
假设有个student类，有id、name、score属性，list集合中存放所有学生信息，现在要根据学生姓名进行分组
10.1 java8之前list分组

public Map<String, List<Student>> groupList(List<Student> students) {
Map<String, List<Student>> map = new Hash<>();
for (Student student : students) {
List<Student> tmpList = map.get(student.getName());
if (tmpList == null) {
tmpList = new ArrayList<>();
tmpList.add(student);
map.put(student.getName(), tmpList);
} else {
tmpList.add(student);
}
}
return map;
}

10.2 java8的list分组

List转map

public Map<String, List<Student>> groupList(List<Student> students) {
	Map<String, List<Student>> map = students.stream().collect(Collectors.groupingBy(Student::getName));
	return map;
}

查看list中是否有重复值
11.1 将list转为set，通过两个集合的size大小是否相等来判断有无重复元素

public static void main(String[] args) {
List<String> stringList=new ArrayList<>(Arrays.asList("a","a","b","c"));
Set<String> stringSet=new HashSet<>(stringList);
if (stringList.size() == stringSet.size()) {
System.out.println("没有重复元素");
} else {
System.out.println("有重复元素");
}
}

11.2 使用jdk8的stream判断

public static void main(String[] args) {
    List<String> stringList=new ArrayList<>(Arrays.asList("a","a","b","c"));
    long count = stringList.stream().distinct().count();
    if (stringList.size() == count) {
    System.out.println("没有重复元素");
    } else {
    System.out.println("有重复元素");
    }
}

11.3 判断一个对象集合中的某个字段是否有重复

判断Person集合中的name属性有没有重复

public static void main(String[] args) {
    List<Person> personList = new ArrayList<Person>(){{
        add(new Person("张三"));
        add(new Person("李四"));
        add(new Person("张三"));
    }};
    List<String> stringList = personList.stream().map(Person::getName)
        .collect(Collectors.toList());
    long count = stringList.stream().distinct().count();
    if (stringList.size() == count) {
        System.out.println("没有重复元素");
    } else {
        System.out.println("有重复元素");
    }
}

判断list集合中是否包含某个元素 list.contains()

List<String> list = new ArrayList<>();
list.add("1");
list.add("2");
list.add("3");
list.add("4");
list.add("5");
boolean isHave = list.contains("2");
list去重

// 去重并转类型
List<String> collect = longList.stream().distinct().map(String::valueOf).collect(Collectors.toList());
从List中取两个字段作为map的key和value

Map<String, Long> collect = studentList.stream().collect(Collectors.toMap(Student::getName, Student::getId));
从list中取一个字段进行分组转map 并过滤

// 并过滤名字为空的数据
HashMap<String,Student> studentMap = studentList.filter(item -> StringUtils.isNotBlank(item.getName())).collect(Collectors.groupingBy(Student::getName));
list倒排序

Collections.reverse(list);