785. Is Graph Bipartite? ----M !!!! !!!!!!!!!

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].

A:

class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        int n = graph.size();
        vector<int> color(n,-1); // 0 as one color, 1 as another, and -1 as no-decided        
        for(int i =0;i<n;i++){
            if(color[i] >= 0) // already visited
                continue;
            color[i] = 1; // found a new connected component
            queue<int> Q;
            Q.push(i);
            while(!Q.empty()){
                int start = Q.front();
                Q.pop();
                for(auto end : graph[start]){
                    if(color[end] == color[start])
                        return false;
                    else if(color[end]== -1){
                        color[end] = 1 - color[start];
                        Q.push(end);
                    }
                }
            }
        }
        return true;
    }
};

注意QUEUE的应用。 （我一直以来都喜欢用2个 layer 但是那样需要拷贝一下，又空间的损失)