java - 如何使用java中的列表结构确定java中当前元素之后是否存在另一个元素

Question

我正在尝试编写一个名为 boolean hasNext() 的函数，它检查当前元素之后是否存在另一个元素 我有一个名为 TourElement 的类，它包含很多点。这是我的代码//类路径点:

public class Waypoint {
    int x  ;
    int y  ;
    public int getX()
    {
        return this.x;
    }
    public int getY()
    {
        return this.y;
    }
    public void setXY(int x, int y)
    {
        this.x = x;
        this.y = y;
    }

//类tourElement

 public class TourElement {
     private Waypoint points;
     private TourElement next;

      public void setWaypoint( Waypoint points){
       this.points = points; 
     }
      public void setTourElement(TourElement next) {
          this.next = next;
      }
     Waypoint getWaypoint() {
         return this.points;
     }

     TourElement getNext(){
         return this.next;
     }

    boolean hasNext(Waypoint first){
    // What am I doing wrong here?
        TourElement current = getNext();
        while( current.next != null)
        {
            return true;
        }
        return false;

    }
    // my test case
         public void testHasNext()
        {
           TourElement elem = createElementList(new int[][] {{0, 0}, {1, 1}, {2, 2}});

            assertEquals(true,elem.hasNext(createWaypoint(1, 1)));
        }

//创建元素列表:

private TourElement createElementList(int[][] waypoints){
        assert waypoints.length > 0;
        TourElement elem = new TourElement();
        int lastIndex = waypoints.length-1;
        Waypoint wp = createWaypoint(waypoints[lastIndex][0], waypoints[lastIndex][1]);
        elem.setWaypoint(wp);
        for (int i = lastIndex-1; i >= 0 ; i--) {
            wp = createWaypoint(waypoints[i][0], waypoints[i][1]);
            elem = elem.addStart(wp);
        }
        return elem;
    }

//创建航路点:

private Waypoint createWaypoint(int x, int y) {
        Waypoint wp = new Waypoint();
        wp.setXY(x, y);
        return wp;
    }

我希望使用 hasNext 函数，如果我传递像 {1,1} 这样的点，它将返回 true，因为在该点之后还有一个点。但是当我通过{2,2}时。它将返回 false

Answer 1

你的方法可以改进。看看吧

boolean hasNext(){
    if (this.next != null) return true;
    return false;
}

您不需要循环到 LinkledList 的末尾，您应该只关心接下来的内容。