Question

请帮我找一个解决方案，以循环/负载均衡的方式从多个请求（UserRequest（userId，timestamp））生成排序LinkedList<UserRequest>？假设我们有多个用户的多个请求列表（XY，其中X：userId，Y：timestamp）： A1，A2，A3，B1，C4，C6 .... 如何将其分类为：
A1，B1，C4，（所有用户最早的要求）
A2，C6，（所有用户的新要求）
A3 ??? （来自所有用户的最新请求）

我已尝试使用LinkedHashMap<userid, LinkedList<timestamp>> 但是现在几天都没有达到效率：

@Test
public void testRequestSorting() throws Exception {
    Collection<UserRequest> userRequests = new LinkedList<>(Arrays.asList(
            new UserRequest("A", 1),
            new UserRequest("A", 2),
            new UserRequest("A", 3),
            new UserRequest("B", 1),
            new UserRequest("C", 4),
            new UserRequest("C", 6)
    ));

    Map<String, LinkedList<Long>> groups = new LinkedHashMap<>();
    for (UserRequest userRequest : userRequests) {
        LinkedList<Long> group = groups.get(userRequest.getUserId());
        if (group == null) {
            group = new LinkedList<>();
            groups.put(userRequest.getUserId(), group);
        }
        group.add(userRequest.getTimestamp());
    }

    LinkedList<UserRequest> sortedRequests = new LinkedList<>();

    while (!groups.isEmpty()) {
        for (Iterator<String> iterator = groups.keySet().iterator(); iterator.hasNext();){
            String userId = iterator.next();
            LinkedList<Long> timestamps = groups.get(userId);
            if (!timestamps.isEmpty()) {
                Long first = timestamps.getFirst();
                sortedRequests.add(new UserRequest(userId, first));
                timestamps.remove(0);
                if (timestamps.isEmpty()) {
                    iterator.remove();
                }
            }
        }
    }

    for (UserRequest sortedRequest : sortedRequests) {
        System.out.println(sortedRequest.getUserId() + "->" + sortedRequest.getTimestamp());
    }
}

class UserRequest {
    private final String userId;
    private final long timestamp;

    public UserRequest(String userId, long timestamp) {
        this.userId = userId;
        this.timestamp = timestamp;
    }

    public String getUserId() {
        return userId;
    }

    public long getTimestamp() {
        return timestamp;
    }
}

循环时尚整理

0 个答案: