785. Is Graph Bipartite?

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题目

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.

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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}.

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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].

解题思路

  • 本题给了无向图的边，判断这个图是否是一个二分图
  • 什么是二分图？

    二分图又称作二部图，是图论中的一种特殊模型。 设G=(V,E)是一个无向图，如果顶点V可分割为两个互不相交的子集(A,B)，并且图中的每条边（i，j）所关联的两个顶点i和j分别属于这两个不同的顶点集(i in A,j in B)，则称图G为一个二分图。孤立点可以任意划分在A或B集合。

  • 解法：
    • 使用BFS + 染色法
    • BFS以未被染色且非孤立点的点作为BFS搜索的起点
    • 使用染色法，只使用两种颜色，将与该点相邻（存在一条边）且未被染色的点染成与其不同的颜色，假如相邻的点已被染色且颜色与该点相同，则证明该图不是二分图。

实现代码

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// 染色法 + BFS

class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        int num = graph.size();
        
        bool colors[num] = {false};
        int visited[num] = {0};

        queue<int> q;
        
        for(int i = 0; i < num; ++i) {
            if(graph[i].size() > 0 && visited[i] == 0) {
                q.push(i);
                visited[i] = 1;
                
                // bfs
                while(!q.empty()) {
                    int top = q.front();
                    q.pop();
                    for(auto i : graph[top]) {
                        if(!visited[i]) {
                            visited[i] = 1;
                            colors[i] = !colors[top];
                            q.push(i);
                        }
                        if(colors[i] == colors[top])
                            return false;
                    }
                }
            }
        }

        return true;
    }
};