Six Degrees

Six degrees of separation is the theory that everyone and everything is six or fewer steps away, by way of introduction, from any other person in the world, so that a chain of "a friend of a friend" statements can be made to connect any two people in a maximum of six steps.

Given a friendship relations, find the degrees of two people, return -1 if they can not been connected by friends of friends.

Example
Gien a graph:

1------2-----4
  \             /
    \         /
     \--3--/
{1,2,3#2,1,4#3,1,4#4,2,3} and s = 1, t = 4 return 2

Gien a graph:

1 2-----4
          /
        /
      3
{1#2,4#3,4#4,2,3} and s = 1, t = 4 return -1

这道题的degreeFromSource不仅用来记录节点到source的距离，同时也可以用来记录是否访问。因为无向图里计算某个节点到source的距离只需要访问一次，不能重复计算。

/**
 * Definition for Undirected graph.
 * class UndirectedGraphNode {
 *     int label;
 *     List<UndirectedGraphNode> neighbors;
 *     UndirectedGraphNode(int x) { 
 *         label = x;
 *         neighbors = new ArrayList<UndirectedGraphNode>(); 
 *     }
 * };
 */
public class Solution {
    /**
     * @param graph a list of Undirected graph node
     * @param s, t two Undirected graph nodes
     * @return an integer
     */
    public int sixDegrees(List<UndirectedGraphNode> graph,
                          UndirectedGraphNode s,
                          UndirectedGraphNode t) {
        // Write your code here
        if (graph == null || s == t || graph.size() == 0){
            return 0;
        }
        Map<UndirectedGraphNode, Integer> degreeFromSource = new HashMap<>();
        Queue<UndirectedGraphNode> queue = new LinkedList<>();
        degreeFromSource.put(s,0);
        queue.offer(s);
        while (!queue.isEmpty()){
            UndirectedGraphNode curt = queue.poll();
            for (UndirectedGraphNode nei : curt.neighbors){
                if (degreeFromSource.containsKey(nei)){
                    continue;
                }
                degreeFromSource.put(nei, degreeFromSource.get(curt) + 1);
                if (nei == t){
                    return degreeFromSource.get(curt) + 1;
                }
                queue.offer(nei);
            }
        }
        return -1;
    }
}