LF.56.Bipartite

/*
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.

Examples

1  --   2

    /

3  --   4

is bipartite (1, 3 in group 1 and 2, 4 in group 2).

1  --   2

    /   |

3  --   4

is not bipartite.

Assumptions

The graph is represented by a list of nodes, and the list of nodes is not null.

*/
/**
 * public class GraphNode {
 *   public int key;
 *   public List<GraphNode> neighbors;
 *   public GraphNode(int key) {
 *     this.key = key;
 *     this.neighbors = new ArrayList<GraphNode>();
 *   }
 * }
 */

//node 和 他的邻居 不能是一样的颜色
public class Solution {
  public boolean isBipartite(List<GraphNode> graph) {
    // write your solution here
    if (graph == null) {
        return true ;
    }
    Map<GraphNode, Integer> visited = new HashMap<>() ;
    for ( GraphNode node : graph ) {
        if (!helper(node, visited)) {
            return false;
        }
    }
    return true;
  }
  private boolean helper(GraphNode node, HashMap<GraphNode, Integer> visited){
    //this node already be handled
    if (visited.containsKey(node)) {
        return true ;
    }
    Queue<GraphNode> queue = new LinkedList<>() ;
    queue.offer(node);
    //assign value
    visited.put(node, 0);
    while(!queue.isEmpty()){
        GraphNode curr = queue.poll() ;
        List<GraphNode> neighbors = curr.neighbors ;
        int currColor = visited.get(curr) ;
        int neiColor = currColor == 0 ? 1 :0 ;
        for ( GraphNode nei : neighbors) {
            /*if the nei has not yet been visited, then color it with neiColor,
            mark it as visited and then offer it */
            if (!visited.containsKey(nei)) {
                visited.put(nei, neiColor);
                queue.offer(nei);
            } else{
                if (visited.get(nei) != neiColor) {
                    return false ;
                }
                /*
                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
                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
                */
            }
        }
    }
    return true ;
  }
}

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
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