LeetCode #785 - Is Graph Bipartite?

最新推荐文章于 2022-06-14 17:21:34 发布

原创最新推荐文章于 2022-06-14 17:21:34 发布 · 214 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

625 篇文章

订阅专栏

本文介绍了一种使用递归着色法来判断无向图是否为二分图的方法。通过给未着色的节点上色，并使其相邻节点上相反颜色，可以检查图是否能被分为两个独立子集，每条边连接不同子集的节点。若出现颜色冲突，则表明图不能被二分。

题目描述：

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].

采用着色法，如果一个点没有被着色，可以让它着一种颜色，然后让它的领接点都着相反的颜色，不断调用递归，可以让一整块连通区域全部着色，一旦有冲突就说明不能二分。

class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        vector<int> colors(graph.size(),0); // 表示每个点的着色情况
        for(int i=0;i<graph.size();i++)
        {
            if(colors[i]==0&&helper(graph,i,1,colors)==false) // 如果i没有被着色，即它没有和之前的任何点连通，那么可以让它的颜色等于1
                return false;
        }
        return true;
    }
   
 
    // 递归函数，判断给一个点着某一种颜色是否可行
    bool helper(vector<vector<int>>& graph, int i, int color, vector<int>& colors)
    {
        if(colors[i]!=0) return colors[i]==color;
        colors[i]=color;
        for(int j=0;j<graph[i].size();j++)
        {
            if(helper(graph,graph[i][j],-color,colors)==false)
                return false;
        }
        return true;
    }
};