我这里获得一串以逗号相隔的字符串,我最后要生成List
List<Long> list = Arrays.stream(ids.split(","))
.map((str) -> Long.parseLong(str))
.collect(Collectors.toList());
List<Integer> list = Arrays.asList(7, 6, 9, 3, 8, 2, 1);
// 遍历输出符合条件的元素
list.stream().filter(x -> x > 6).forEach(System.out::println);
// 是否包含符合特定条件的元素
boolean anyMatch = list.stream().anyMatch(x -> x > 6);
使用filter可以进行筛选
List<String> list = Arrays.asList("adnm", "admmt", "pot", "xbangd", "weoujgsd");
Optional<String> max = list.stream().max(Comparator.comparing(String::length));
max里边是comparator,比较器,传入一个这个就知道怎么比较辣
List<Integer> list = Arrays.asList(7, 6, 4, 8, 2, 11, 9);
long count = list.stream().filter(x -> x > 6).count();
System.out.println("list中大于6的元素个数:" + count);
统计大于6的个数
String[] strArr = { "abcd", "bcdd", "defde", "fTr" };
List<String> strList = Arrays.stream(strArr).map(String::toUpperCase).collect(Collectors.toList());
List<Integer> intList = Arrays.asList(1, 3, 5, 7, 9, 11);
List<Integer> intListNew = intList.stream().map(x -> x + 3).collect(Collectors.toList());
System.out.println("每个元素大写:" + strList);
System.out.println("每个元素+3:" + intListNew);
归纳操作,一般拿来求和 求乘积,求最值
List<Integer> list = Arrays.asList(1, 3, 2, 8, 11, 4);
// 求和方式1
Optional<Integer> sum = list.stream().reduce((x, y) -> x + y);
// 求和方式2
Optional<Integer> sum2 = list.stream().reduce(Integer::sum);
// 求和方式3
Integer sum3 = list.stream().reduce(0, Integer::sum);
// 求乘积
Optional<Integer> product = list.stream().reduce((x, y) -> x * y);
// 求最大值方式1
Optional<Integer> max = list.stream().reduce((x, y) -> x > y ? x : y);
// 求最大值写法2
Integer max2 = list.stream().reduce(1, Integer::max);
System.out.println("list求和:" + sum.get() + "," + sum2.get() + "," + sum3);
System.out.println("list求积:" + product.get());
System.out.println("list求最大值:" + max.get() + "," + max2);