785. Is Graph Bipartite?

• 題目
  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.

  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].

• 解題思路

  • 本題給了無向圖的邊，判斷這個圖是否是一個二分圖
  • 什麼是二分圖？

    二分圖又稱作二部圖，是圖論中的一種特殊模型。 設G=(V,E)是一個無向圖，如果頂點V可分割為兩個互不相交的子集(A,B)，並且圖中的每條邊（i，j）所關聯的兩個頂點i和j分別屬於這兩個不同的頂點集(i in A,j in B)，則稱圖G為一個二分圖。孤立點可以任意劃分在A或B集合。

  • 解法：
    • 使用BFS + 染色法
    • BFS以未被染色且非孤立點的點作為BFS搜尋的起點
    • 使用染色法，只使用兩種顏色，將與該點相鄰（存在一條邊）且未被染色的點染成與其不同的顏色，假如相鄰的點已被染色且顏色與該點相同，則證明該圖不是二分圖。

• 實現程式碼

  // 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;
      }
  };