我设法使用 Java 8 Streams API 编写了一个解决方案,该解决方案首先按对象 Route 的值对列表进行分组,然后计算每组中的对象数量。它返回一个映射 Route -> Long。这是代码:
Map<Route, Long> routesCounted = routes.stream()
.collect(Collectors.groupingBy(gr -> gr, Collectors.counting()));
和路线类:
public class Route implements Comparable<Route> {
private long lastUpdated;
private Cell startCell;
private Cell endCell;
private int dropOffSize;
public Route(Cell startCell, Cell endCell, long lastUpdated) {
this.startCell = startCell;
this.endCell = endCell;
this.lastUpdated = lastUpdated;
}
public long getLastUpdated() {
return this.lastUpdated;
}
public void setLastUpdated(long lastUpdated) {
this.lastUpdated = lastUpdated;
}
public Cell getStartCell() {
return startCell;
}
public void setStartCell(Cell startCell) {
this.startCell = startCell;
}
public Cell getEndCell() {
return endCell;
}
public void setEndCell(Cell endCell) {
this.endCell = endCell;
}
public int getDropOffSize() {
return this.dropOffSize;
}
public void setDropOffSize(int dropOffSize) {
this.dropOffSize = dropOffSize;
}
@Override
/**
* Compute hash code by using Apache Commons Lang HashCodeBuilder.
*/
public int hashCode() {
return new HashCodeBuilder(43, 59)
.append(this.startCell)
.append(this.endCell)
.toHashCode();
}
@Override
/**
* Compute equals by using Apache Commons Lang EqualsBuilder.
*/
public boolean equals(Object obj) {
if (!(obj instanceof Route))
return false;
if (obj == this)
return true;
Route route = (Route) obj;
return new EqualsBuilder()
.append(this.startCell, route.startCell)
.append(this.endCell, route.endCell)
.isEquals();
}
@Override
public int compareTo(Route route) {
if (this.dropOffSize < route.dropOffSize)
return -1;
else if (this.dropOffSize > route.dropOffSize)
return 1;
else {
// if contains drop off timestamps, order by last timestamp in drop off
// the highest timestamp has preceding
if (this.lastUpdated < route.lastUpdated)
return -1;
else if (this.lastUpdated > route.lastUpdated)
return 1;
else
return 0;
}
}
}
我还想实现的是,每个组的键都是具有最大 lastUpdated 值的组。我已经在看这个解决方案 https://stackoverflow.com/a/24274203/535661但我不知道如何将计数和分组按值与路由最大lastUpdated值结合起来。这是我想要实现的示例数据:
EXAMPLE:
List<Route> routes = new ArrayList<>();
routes.add(new Route(new Cell(1, 2), new Cell(2, 1), 1200L));
routes.add(new Route(new Cell(3, 2), new Cell(2, 5), 1800L));
routes.add(new Route(new Cell(1, 2), new Cell(2, 1), 1700L));
应转换为:
Map<Route, Long> routesCounted = new HashMap<>();
routesCounted.put(new Route(new Cell(1, 2), new Cell(2, 1), 1700L), 2);
routesCounted.put(new Route(new Cell(3, 2), new Cell(2, 5), 1800L), 1);
请注意,映射的键(计算 2 个路由)是第一个具有最大的lastUpdated值.