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 containi
or duplicate values.- The graph is undirected: if any element
j
is ingraph[i]
, theni
will be ingraph[j]
.
class Solution { public: bool dfs(vector<vector<int> >& graph, vector<int>& state, int i, int color){ for (int j = 0; j<graph[i].size(); j++){ if (state[graph[i][j]] == 0){ //没有遍历到时 state[graph[i][j]] = -color; //标记该节点颜色同时继续搜索 return dfs(graph, state, graph[i][j], -color); } else if (state[graph[i][j]] == color){ //邻居节点中与该节点颜色相同则返回false return false; } } return true; } bool isBipartite(vector<vector<int>>& graph) { int node_num = graph.size(); vector<int> state(node_num,0); int result = true; for(int i=0; i<graph.size(); i++){ if(state[i]==0 && !dfs(graph, state, i, 1)) result = false; } return result; } };