你可以用 Collectors.groupingBy .

Map<Integer, List<Account>> map =
    accounts.stream().collect(Collectors.groupingBy(Account::getCustomerID));

演示

List<Account> accounts = new ArrayList<>();
accounts.add(new Account(1, "abc", 17998210, 190));
accounts.add(new Account(2, "hsj", 6786179, 190));
accounts.add(new Account(4, "ioip", 246179, 191));
accounts.add(new Account(4, "ewrew", 90179, 191));

Map<Integer, List<Account>> accountsMap = new HashMap<>();

for (Account account : accounts) {
    accountsMap.computeIfAbsent(account.customerID, k -> new ArrayList<>()).add(account);
}

我宁愿使用 Collectors.toMap 方法具有三个参数，以确保清晰：

Map<Integer, List<Account>> map = accounts.stream()
        .collect(Collectors.toMap(
                // key - customerID
                e -> e.getCustomer().getCustomerID(),
                // value - List<Account>
                e -> List.of(e),
                // merge two lists
                (l1, l2) -> Stream.of(l1, l2)
                        .flatMap(List::stream)
                        .collect(Collectors.toList())));