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

Example 2: Input: [[1,2,3], [0,2], [0,1,3], [0,2]] Output: false

解題思路

本題是判斷一個無向圖是否為二分圖，我使用染色法進行相鄰節點的判斷，並採取BFS進行圖的遍歷

1. 初始化儲存顏色的陣列 color（0表示白色，1表示黑色，-1表示未染色），建立BFS佇列以及判斷該節點是否被遍歷過的狀態陣列 check
2. 從任意節點開始進行遍歷，將該節點新增到BFS佇列，顏色設為1，check 狀態為true（我選擇了0號節點）
3. 每次從BFS佇列中取出一個新節點，若該節點的子節點未被訪問過（染色），那麼將其染成與當前節點相反的顏色，並將這個子節點新增到BFS佇列的尾部，設定 check
   狀態為true；否則，檢查該子節點的顏色與當前節點是否相同，若相同，說明圖中出現了奇數邊的環，那麼這個圖不是二分圖，返回false，結束
4. 由於測試樣例中有無出度的節點，因此在BFS佇列為空的時候要檢查一下狀態陣列 check 中是否還有未遍歷的節點；若有，將其新增到BFS佇列，繼續遍歷
5. 當所有節點已經訪問完畢，沒有出現顏色重複的節點，說明這個圖是二分圖，返回true

C++程式碼

class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        vector<
 
int> color(graph.size(), -1);
        queue<int> bfs_queue;
        vector<bool> check(graph.size(), false);
        
        color[0] = 1;
        bfs_queue.push(0);
        check[0] = true;
        
        while(!bfs_queue.empty()) {
            int p = bfs_queue.front();
            bfs_queue.pop();
            for(int i = 0; i < graph[p].size(); i++){
                if(color[graph[p][i]] == -1){
                    color[graph[p][i]] = color[p] == 1? 0 : 1;
                    bfs_queue.push(graph[p][i]);
                    check[graph[p][i]] = true;
                }
                if(color[graph[p][i]] == color[p])
                    return false;
            }
            
            if(bfs_queue.empty()) {
                for(int i = 0; i < check.size(); i++) {
                    if(!check[i]) {
                        bfs_queue.push(i);
                        color[i] = 1;
                        check[i] = true;
                        break;
                    }
                }
            }
        }
        return true;
    }
};