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

There is an undirected graph with n nodes, where each node is numbered between 0 and n - 1. You are given a 2D array graph, where graph[u] is an array of nodes that node u is adjacent to. More formally, for each v in graph[u], there is an undirected edge between node u and node v. The graph has the following properties:

• There are no self-edges (graph[u] does not contain u).
• There are no parallel edges (graph[u] does not contain duplicate values).
• If v is in graph[u], then u is in graph[v] (the graph is undirected).
• The graph may not be connected, meaning there may be two nodes u and v such that there is no path between them.

A graph is bipartite if the nodes can be partitioned into two independent sets A and B such that every edge in the graph connects a node in set A and a node in set B.

Return true if and only if it is bipartite.

 

Example 1:

Input: graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
Output: false
Explanation: There is no way to partition the nodes into two independent sets such that every edge connects a node in one and a node in the other.

Example 2:

Input: graph = [[1,3],[0,2],[1,3],[0,2]]
Output: true
Explanation: We can partition the nodes into two sets: {0, 2} and {1, 3}.

 

Constraints:

• graph.length == n
• 1 <= n <= 100
• 0 <= graph[u].length < n
• 0 <= graph[u][i] <= n - 1
• graph[u] does not contain u.
• All the values of graph[u] are unique.
• If graph[u] contains v, then graph[v] contains u.

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)

class Solution {

public:

bool isBipartite(vector<vector<int>>& graph) {

int n = graph.size(); // color the nodes

vector<int> Color(n, -1); // 0(white). and 1 for black

for (int i = 0; i < n; i++) {

int nextColor;

if(Color[i] < 0){

Color[i] = 0;

nextColor = 1;

}else{

nextColor = 1-Color[i];

}

for(auto id2 : graph[i]){

if(Color[id2] < 0 || Color[id2] == nextColor){

Color[id2] = nextColor;

}else{

return false;

}

}

}

return true;

}

};

上面这样是不对的。 自己考虑下，为什么。 （可以跟上上次的解法对比）