785. Is Graph Bipartite?

Probelm:

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.

 Analysis:

这种太数学的题最恶心，没刷过100%做不出来，遇到只有认栽。主要就是标记颜色，一开始让所有node颜色都为0，然后用BFS或者DFS遍历。把edge两端的node标记为不同颜色1, -1。如果遇到edge两边的node颜色一样则说明不是bipartite.

Solution:

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class Solution { public boolean isBipartite(int[][] graph) { Queue<Integer> queue = new LinkedList<>(); queue.offer(0); int[] color = new int[graph.length]; for (int i = 0; i < graph.length; i++) { if (color[i] != 0) continue; queue.offer(i); color[i] = 1; while (!queue.isEmpty()) { int cur = queue.poll(); for (int neib: graph[cur]) { if (color[neib] == 0) { color[neib] = -1 * color[cur]; queue.offer(neib); } else { if (color[neib] == color[cur]) return false; } } } } return true; } }