java.util.HashMap.get方法为一个键返回2个值

时间:2019-02-09 11:49:28

标签: java graph get hashmap

我制作了一个Graph类,该类使用HashMap>将节点保留为键,并将相应边保留为值。

public class GraphAL {
private HashMap<Integer, ArrayList<Integer>> adjList;
private ArrayList<Integer> vertices;
int numberOfNodes;
boolean visited[];

GraphAL(HashMap<Integer, ArrayList<Integer>> adjList, 
        ArrayList<Integer> vertices, int numberOfNodes){
    this.adjList = adjList;
    this.vertices = vertices;
    this.numberOfNodes = numberOfNodes;
    visited = new boolean[this.numberOfNodes];
}

public HashMap<Integer, ArrayList<Integer>>  getAdjList() {
    return adjList;
}

public ArrayList<Integer>  getVertices() {
    return vertices;
}

public int getNumeberOfNodes()
{
    return numberOfNodes;
}

public boolean[] getVIsitedNodes()
{
    return visited;
}

public void setVisitedNodesToTrue(int node)
{
    visited[node] = true;
}

}

然后,我制作了一种反转图形的方法。 问题是尽管我的键是唯一的,但是java.util.HashMap.get方法返回2个值而不是1个。这导致将边缘添加到的位置比我要添加边缘的确切节点更多。

public static GraphAL reverseGraph(GraphAL g)
{
    HashMap<Integer, ArrayList<Integer>> revAdjList = new HashMap<Integer, ArrayList<Integer>>();

    ArrayList<Integer> revVertices = new ArrayList<Integer>();

    System.out.println("Printing in for loop");
    for(Integer x:g.getVertices())
    {
        System.out.println("x = " + x);
        ArrayList<Integer> edges = new ArrayList<Integer>();
        edges.add(x);
        System.out.println("Edges: ");
        for(Integer m:edges)
        {
        System.out.print(m +  " ");
        }
        for(Integer y:g.getAdjList().get(x))
        {
            System.out.println("y = " + y);

            if(!revVertices.contains(y)) 
            {
                    revVertices.add(y);
                    System.out.println("RevVertices: ");
                    for(Integer n:revVertices)
                    {
                    System.out.print(n + " ");
                    }
            }
            if(revAdjList.containsKey(y))
            {
                for(Integer o:revAdjList.get(5))
                {
                    System.out.println(5 + " contains " + o);
                }
                System.out.println("adding " + x + " to " + y);

                revAdjList.get(y).add(x);

                for(Integer o:revAdjList.get(y))
                {
                    System.out.println(y + " contains " + o);
                }

                for(Integer o:revAdjList.get(5))
                {
                    System.out.println(5 + " contains " + o);
                }

            }
            else
            {
            revAdjList.put(y, edges);
            System.out.println("putting " + x + " at " + y);
            }
            System.out.println("Current AdjList: ");
            for(Integer h:revVertices)
            {
                System.out.print("vertice is: " + h + " ");
                for(Integer j:revAdjList.get(h))
                {
                    System.out.print("edge : " + j + " ");

                }
                System.out.println();
            }
        }

    }
    System.out.println("Done printing in for loop");

    GraphAL revGraph = new GraphAL(revAdjList, revVertices, g.getNumeberOfNodes());
    return revGraph;
}

对所有print.ln的文档表示抱歉,但我想确定问题出在哪里。它看起来像revAdjList.get(y).add(x),其中y = 6和x = 3像我想要的那样也从键6和键5返回对应的ArrayList。Ofc这导致将边3添加到6就像我想要的一样,但它也将其添加到5。 有想法吗?

x = 3
Edges: 
3 y = 6
5 contains 8
adding 3 to 6
6 contains 8
6 contains 3
5 contains 8
5 contains 3
adding 3 to 6
Current AdjList: 
vertice is: 1 edge : 7 
vertice is: 2 edge : 5 
vertice is: 3 edge : 9 
vertice is: 7 edge : 9 
vertice is: 4 edge : 1 
vertice is: 5 edge : 8 edge : 3 
vertice is: 6 edge : 8 edge : 3 

1 个答案:

答案 0 :(得分:0)

问题在这条线上发生

for (Integer j : revAdjList.get(h)) {
    System.out.print("edge : " + j + " ");
}

让我们在此指出问题以及为什么两次打印"edge"

revAdjList.get(h)

返回Integer的集合,而不是单个Integer,因此它可以具有任意数量的整数值,因为您将其定义为集合而不是单吨。

您要从此处添加多个边缘

for (Integer m : edges) {
    System.out.print(m + " ");
}

如果只需要1条边,请不要将edges做成数组。

HashMap<Integer, ArrayList<Integer>> revAdjList = new HashMap<>();

应定义为

HashMap<Integer, Integer> revAdjList = new HashMap<>();

完成此操作后,您可以删除循环并清理代码以使用新语法。