如何找到入射到特定顶点的边列表

Question

I tried the following but I'm not sure that it is correct.

ArrayList<ArrayList<Integer>> list = new ArrayList<>();
public static ArrayList<ArrayList<Integer>> incidentEdges(int v) {
   for(int i = 0; i < a.length; i++) {
      for(int j = 0; j < a[i].length; j++) {
         if(a[v][j] == 1) {
            list.get(v).get(new Edge(start, destination));
            list.get(j).get(new Edge(start, destination);
         }
      }
   }
   return list;
}

The array a is an adjacency matrix and the parameter v is the vertex of an undirected graph. If there is an edge between vertex v and j then we add the edge incident to vertex v .

Answer 1

Method 1: Query Adjacency Matrix

Since you have already stored the edges in an adjacency matrix, you can simply query it. Set your i to v (since you are not even using i in the first place), and then check all vertices that are connected.

public static ArrayList<Integer> incidentEdges(int v) {
   ArrayList<Integer> result = new ArrayList<>();
   for(int i = 0; i < a[v].length; i++) {
     if(a[v][i] == 1) {
        result.add(a[v].get(i));
     }
   }
   return result;
}

Method 2: Generating an Adjacency List

If you get control of the input (ie before making the adjacency matrix), what you want is an adjacency list : Where each list[start] points to an ArrayList<Integer> (representing the connected vertices). I would avoid the use of ArrayList since the number of vertices is already known. So I would instead use ArrayList<Integer>[] list instead. This definitely makes it easier to code.

Here is an example of generating the adjacency list from an adjacency matrix

static ArrayList<Integer>[] list;

public static ArrayList<Integer>[] generateAl(int v) {
   list = new ArrayList[a.length];
   for(int i = 0; i < a.length; i++) {
      list[i] = new ArrayList<>();
      for(int j = 0; j < a[i].length; j++) {
         if(a[i][j] == 1) {
            list[i].add(j);
         }
      }
   }
   return list;
}

public static ArrayList<Integer> incidentEdges(int v) {
   return list[v];
}