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.

Solution: 對每個節點染色，如果相鄰兩個節點顏色相同，則不能。反之，則能。

class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        int color[graph.size()];
        int visitedNode[graph.size()];
        int size,
 
 index;
        for (int i=0; i<graph.size(); ++i) {
            color[i] = -1;
        }
        for (int i=0; i<graph.size(); ++i) {
            if (color[i] == -1) {
                visitedNode[0] = i;
                color[i] = 1;
                index = 0;
                size = 1;
                while (index < size) {
                    int currentNode = visitedNode[index];
                    ++index;
                    for (int j=0; j<graph[currentNode].size(); ++j) {
                        if (color[graph[currentNode][j]] == -1) {
                            visitedNode[size] = graph[currentNode][j];
                            ++size;
                            color[graph[currentNode][j]] = !color[currentNode];
                        } else if (color[graph[currentNode][j]] == color[currentNode]) {
                            return false;
                        }
                    }
                }
            }
        }
        return true;
    }
};