Is Graph Bipartite?

LeetCode

第21天。

今天的题目是 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.

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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}.
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].

这是一道关于图的问题，题目的意思很简单，就是要判断一个图是不是一个二部图，所谓的二部图，就是一个图可以把所有节点划分到两个不相交的两个集合，这两个集合内部没有边相连。

我们可以对图进行一次遍历，遍历的时候对节点进行着色，着色的规律是这样的，当从一个节点跳到另一个节点的时候，我们就切换一次颜色（共有三种颜色，其中一种表示没有访问，即白色）。因为遍历完了之后，整个图的节点就被划分成两部分了，接下来我们只需要判断所有节点的邻居是否和它是不同色的即可。代码如下：

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char color;
bool isBipartite(vector<vector<int>>& graph) {
    int size = graph.size();
    vector<char> flags(size, 'w');
    // w g b
    color = 'b';
    for(int i = 0;i < size;i++) {
        if (flags[i] == 'w') {
            dfs(graph, flags, i);
        }
    }

    for(int i = 0;i < size;i++) {
        for(auto j: graph[i]) {
            if (flags[i] == flags[j]) return false;
        }
    }
    return true;
}

void dfs(vector<vector<int>> &graph, vector<char> &flags, int index) {
    flags[index] = color;
    for(auto j: graph[index]) {
        if (flags[j] == 'w') {
            color = ~color;
            dfs(graph, flags, j);
            color = ~color;
        }
    }
}