lintcode531- Six Degrees- medium- microsoft

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

 

层级遍历的BFS。queue + set + size管层 + steps计数。1.入第一个人  2.while循环每一层  3.for循环当前层内每个点，如果找到第二个人了，返回计数器。然后做检测邻居加邻居处理。

 

/**
 * 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: Undirected graph node
     * @param t: Undirected graph nodes
     * @return: an integer
     */
    public int sixDegrees(List<UndirectedGraphNode> graph, UndirectedGraphNode s, UndirectedGraphNode t) {
        // write your code here
        
        if (graph == null || graph.size() == 0 || s == null || t == null) {
            return -1;
        }
        
        Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
        Set<UndirectedGraphNode> set = new HashSet<UndirectedGraphNode>();
        queue.offer(s);
        set.add(s);
        int steps = 0;
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                UndirectedGraphNode node = queue.poll();
                if (node == t) {
                    return steps;
                }
                for (UndirectedGraphNode neighbor : node.neighbors) {
                    if (!set.contains(neighbor)) {
                        queue.offer(neighbor);
                        set.add(neighbor);
                    }
                }
            }
            steps++;
        }
        
        return -1;
    }
}