LF.56.Bipartite

 1 /*
 2 Determine if an undirected graph is bipartite. A bipartite graph is one in which the nodes 
can be divided into two groups such that no nodes have direct edges to other nodes in the same group.
 3 
 4 Examples
 5 
 6 1  --   2
 7 
 8     /
 9 
10 3  --   4
11 
12 is bipartite (1, 3 in group 1 and 2, 4 in group 2).
13 
14 1  --   2
15 
16     /   |
17 
18 3  --   4
19 
20 is not bipartite.
21 
22 Assumptions
23 
24 The graph is represented by a list of nodes, and the list of nodes is not null.
25 
26 */
27 /**
28  * public class GraphNode {
29  *   public int key;
30  *   public List<GraphNode> neighbors;
31  *   public GraphNode(int key) {
32  *     this.key = key;
33  *     this.neighbors = new ArrayList<GraphNode>();
34  *   }
35  * }
36  */

 1 //node 和 他的邻居 不能是一样的颜色
 2 public class Solution {
 3   public boolean isBipartite(List<GraphNode> graph) {
 4     // write your solution here
 5     if (graph == null) {
 6         return true ;
 7     }
 8     Map<GraphNode, Integer> visited = new HashMap<>() ;
 9     for ( GraphNode node : graph ) {
10         if (!helper(node, visited)) {
11             return false;
12         }
13     }
14     return true;
15   }
16   private boolean helper(GraphNode node, HashMap<GraphNode, Integer> visited){
17     //this node already be handled
18     if (visited.containsKey(node)) {
19         return true ;
20     }
21     Queue<GraphNode> queue = new LinkedList<>() ;
22     queue.offer(node);
23     //assign value
24     visited.put(node, 0);
25     while(!queue.isEmpty()){
26         GraphNode curr = queue.poll() ;
27         List<GraphNode> neighbors = curr.neighbors ;
28         int currColor = visited.get(curr) ;
29         int neiColor = currColor == 0 ? 1 :0 ;
30         for ( GraphNode nei : neighbors) {
31             /*if the nei has not yet been visited, then color it with neiColor,
32             mark it as visited and then offer it */
33             if (!visited.containsKey(nei)) {
34                 visited.put(nei, neiColor);
35                 queue.offer(nei);
36             } else{
37                 if (visited.get(nei) != neiColor) {
38                     return false ;
39                 }
40                 /*
41                 做图的时候千万要小心，如果 当前点的邻居已经访问过了， 则不要再放入queue 中
42                 否则一定死循环 所以图的 bfs 一定要 用一个字典来mark 访问过的点
43                 */
44             }
45         }
46     }
47     return true ;
48   }
49 }