785. 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.

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

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public boolean isBipartite(int[][] graph) {
int[] visited = new int[graph.length];

for(int i = 0; i < graph.length; i++){
if(visited[i] == 0){
Deque<Integer> q = new ArrayDeque<>();
q.offer(i);
visited[i] = 1;
while(!q.isEmpty()){
Integer node = q.poll();
for(int j = 0; j < graph[node].length; j++){
if(visited[graph[node][j]] == 0){
q.offer(graph[node][j]);
visited[graph[node][j]] = visited[node] == 1 ? 2 : 1;
}else{
if(visited[graph[node][j]] == visited[node]) return false;
}
}
}
}
}
return true;
}