如何使用广度优先搜索获得两个节点之间的最短路径？

Question

I am brushing up my algorithm and data structure knowledge and was wondering if someone could help me figure out how to find the shortest path between two nodes with BFS . So far I am have the following method, which returns the shortest parth in a graph:

private Node[] nodes;

public static int[] shortestPath(Node source){
        LinkedList<Node> queue = new LinkedList<Node>();
        queue.add(source);

        int[] distances = new int[totalNodes];
        Arrays.fill(distances,-1);

        distances[source] = 0;

        while(!queue.isEmpty()){
            Node node = queue.poll();

            for (int neighbor: nodes[node].neighbor) {
                if(distances[neighbor] == -1) {
                    distances[neighbor] += distances[distanceIndex] + 1;
                    queue.add(neighbor);
                }

            }
        }
            return distances;
    }


}

I was wondering how would the solution look if I wanted to implement a method like this:

public static int[] shortestPath(Node source, Node destination){
// 
}

Thanks in advance for the help, I am new to data structures and not sure how to go about it

Answer 1

private Node[] nodes;

public static int shortestPath(Node source, Node destination){
        LinkedList<Node> queue = new LinkedList<Node>();
        queue.add(source);

        int[] distances = new int[totalNodes];
        Arrays.fill(distances,-1);

        distances[source] = 0;

        while(!queue.isEmpty()){
            Node node = queue.poll();

            for (int neighbor: nodes[node].neighbor) {
                if(distances[neighbor] == -1) {
                    distances[neighbor] += distances[node] + 1;
                    queue.add(neighbor);

                    if(neighbor == destination)
                      break;
                }

            }
        }
            return distances[destination];
    }