Six Degrees

六度分隔理论实现

最新推荐文章于 2019-08-06 00:14:14 发布

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最新推荐文章于 2019-08-06 00:14:14 发布

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分类专栏： leetcode

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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.

Have you met this question in a real interview?

Yes

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

java

/**
 * 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 0;
        }
        if (s == t) {
            return 0;
        }
        Map<UndirectedGraphNode, Integer> map = new HashMap<>();
        Queue<UndirectedGraphNode> queue = new LinkedList<>();
        map.put(s, 0);
        queue.offer(s);
        while (!queue.isEmpty()) {
            UndirectedGraphNode node = queue.poll();
            int step = map.get(node);
            for (UndirectedGraphNode val : node.neighbors) {
                if (val == t) {
                    return step + 1;
                }
                if (!map.containsKey(val)) {
                    map.put(val, step + 1);
                    queue.offer(val);
                }
            }
        }
        return -1;
    }
}