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.

Note:

graph will have length in range [1, 100].

graph[i] will contain integers in range [0, graph.length - 1].

graph[i] will not contain i or duplicate values.

The graph is undirected: if any element j is in graph[i], then i will be in graph[j].

## Solution

采用BFS涂色法

BFS树第0层涂红色，第1层涂蓝色，第2层涂红色 。。。 以此类推

检查每条边如果存在一条边两遍涂的颜色相同，则不是二分图。

注意点：图中可能存在多个连通分量

## Code

1 | public boolean isBipartite(int[][] graph) { |