785. Is Graph Bipartite?

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].

--------------------------
垂直一条线把graph分为2半，每一半的node只能有链接通向对面一半，不能跟相同一半的node有链接。

把2半分别标记别true跟false

Solution #1 DFS

class Solution {
    public boolean isBipartite(int[][] graph) {
        Map<Integer, Boolean> map = new HashMap<>();
        for (int i = 0; i < graph.length; i++) {
            if (!map.containsKey(i) && !dfs(map, graph, i, true)) return false;
        }
        
        return true;
    }
    
    private boolean dfs(Map<Integer, Boolean> map, int[][] graph, int node, boolean flag) {
        if (map.containsKey(node) && map.get(node) == flag) return false;
        if (map.containsKey(node)) return true;
        
        map.put(node, !flag);
        for (int neighbor : graph[node]) {
            if (!dfs(map, graph, neighbor, !flag)) return false;
        }
        
        return true;
    }
}

Solution #2 BFS

class Solution {
    
    public boolean isBipartite(int[][] graph) {
        Map map = new HashMap<>();
        for (int i = 0; i < graph.length; i++) {
            if (!map.containsKey(i) && !bfs(map, graph, i)) return false;
        }
        
        return true;
    }
    
    private boolean bfs(Map map, int[][] graph, int node) {
        
        Queue que = new LinkedList<>();
        
        que.add(node);
        map.put(node, true);
        
        while (!que.isEmpty()) {
            int top = que.poll();
            for (int neighbor : graph[top]) {
                if (!map.containsKey(neighbor)) {
                    que.add(neighbor);
                    map.put(neighbor, !map.get(top));
                }else if (map.containsKey(neighbor) && map.get(neighbor) == map.get(top)) {
                    return false;
                }
            }
        }
        
        return true;
    }
    
}