Is Graph Bipartite

Given an undirected graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists.  Each node is an integer between 0 and graph.length - 1.  There are no self edges or parallel edges: graph[i] does not contain i, and it doesn't contain any element twice.

Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation: 
The graph looks like this:
0----1
|    |
|    |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.

Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation: 
The graph looks like this:
0----1
| \  |
|  \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.

思路：bipartite，就是用两种color标记点，相邻的点，不能是一种颜色，题目给的graph刚好是一个adject list

这里注意一点，不能用一个node开始做queue search,因为graph可能有好几个connected component ，所以必须每个点来判断，所以这题用in[] color， 1, -1来表示不同的color，这样就可以判断，所以queue的while循环要写在for(int i = 0; i < graph.length; i++)里面；

class Solution {
    public boolean isBipartite(int[][] graph) {
        if(graph == null ) {
            return false;
        }
        
        // 0 uncolor;
        // 1 one color
        // -1 the other color;
        int[] color = new int[graph.length];
        Queue<Integer> queue = new LinkedList<Integer>();
        for(int i = 0; i < graph.length; i++) {
            if(graph[i] != null && color[i] == 0) {
                queue.offer(i);
                color[i] = 1;
            }
            while(!queue.isEmpty()) {
                int size = queue.size();
                for(int j = 0; j < size; j++) {
                    Integer node = queue.poll();
                    for(Integer neighbor: graph[node]) {
                        if(color[neighbor] == color[node]) {
                            return false;
                        }
                        if(color[neighbor] == 0) {
                            color[neighbor] = -color[node];
                            queue.offer(neighbor);
                        }
                    }
                }
            }
        }
        return true;
    }
}