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].

class Solution {
    public boolean isBipartite(int[][] graph) {
        //BFS
        // 0(not meet), 1(black), 2(white)
        int[] visited = new int[graph.length];
        
        for (int i = 0; i < graph.length; i++) {
            if (graph[i].length != 0 && visited[i] == 0) {
                visited[i] = 1;
                Queue<Integer> q = new LinkedList<>();
                q.offer(i);
                while(! q.isEmpty()) {
                    int current = q.poll();
                    for (int c: graph[current]) {

                            if (visited[c] == 0) {
                                visited[c] = (visited[current] == 1) ? 2 : 1;
                                q.offer(c);
                            } else {
                                if (visited[c] == visited[current]) return false;
                            }
                    }
                }                                       
            }
        }        
        return true;
    }
}

BFS，涂色法，对每个点涂色，然后对current它所有的adjacent node涂成相反颜色，只要有一步的检查里和current颜色相同就返回false

 public boolean isBipartite(int[][] g) {
        int[] colors = new int[g.length];
        for (int i = 0; i < g.length; i++)
            if (colors[i] == 0 && !validColor(g, colors, 1, i))
                return false;
        return true;
    }

    boolean validColor(int[][] g, int[] colors, int color, int node) {
        if (colors[node] != 0)
            return colors[node] == color;
        colors[node] = color;
        for (int adjacent : g[node])
            if (!validColor(g, colors, -color, adjacent))
                return false;
        return true;
    }

DFS

https://www.youtube.com/watch?v=670Gn4e89B8