LeetCode 785. Is Graph Bipartite?

問題描述：

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]

輸入輸出示例：

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].

題解：

判斷一個圖是否是二部圖，最常見的方法就是染色法，即先選取一個點染為任意顏色，然後遍歷這個點的邊，對其鄰點進行染色（與這個點不同的顏色），如果發現其鄰點已染色，則有兩種情況：

1. 與該點同色，則說明該圖不是二部圖
2. 與該點異色，則說明之前其鄰點已經遍歷過了

Ok，想法很簡單，用 BFS 或 DFS 都能簡單地完成這個任務，只是需要注意這個圖可能不是一個連通圖，所以應該從每個節點開始遍歷，避免漏掉某些節點。

Code：

/*
 * 0 代表未染色，1 和 -1 表示兩種不同的顏色
 */
bool isBipartite(vector<vector<int> >& graph) {
	vector<int> colorOfNode(graph.size(), 0);
	
	for (int k = 0; k < graph.size(); ++k) {

		queue<int> openList;
		if (colorOfNode[k] == 0) openList.push(k);

		while (!openList.empty()) {
			int u = openList.front();
			openList.pop();
			if (colorOfNode[u] == 0) colorOfNode[u] = -1;

			for (int i = 0; i < graph[u].size(); ++i) {
				int v = graph[u][i];
				if (colorOfNode[v] == colorOfNode[u]) return false;
				else if (colorOfNode[v] == -colorOfNode[u]) continue;
				else {
					colorOfNode[v] = -colorOfNode[u];
					openList.push(v);
				}
			}

		}
	}
	return true;
}

複雜度分析：

每個節點只會被遍歷一次，然後與節點相連的邊大部分情況下都會被遍歷一次（這裡指 u->v ，因為這個圖本身是個無向圖），所以複雜度是 O(|V|+|E|)。所以這是一個線性判斷一個圖是否為二部圖的演算法。