Code & Fun

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;
}
}
}

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