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.
M1: BFS
time: O(V + E), space: O(V)
class Solution { public boolean isBipartite(int[][] graph) { int[] visited = new int[graph.length]; for(int i = 0; i < graph.length; i++) { if(!BFS(graph, i, visited)) { return false; } } return true; } private boolean BFS(int[][] graph, int node, int[] visited) { if(visited[node] != 0) { return true; } Queue<Integer> q = new LinkedList<>(); q.offer(node); visited[node] = 1; while(!q.isEmpty()) { int cur = q.poll(); int curColor = visited[cur]; int neiColor = -curColor; for(int nei : graph[cur]) { if(visited[nei] == 0) { visited[nei] = neiColor; q.offer(nei); } else if(visited[nei] != neiColor) { return false; } } } return true; } }
M2: DFS
Graph Coloring (by DFS/BFS): color a node as red, and then color all its neighbors as blue recursively. if there's any conflict, return false.
染色法 Graph Coloring (by DFS)
将相连的两个顶点染成不同的颜色，一旦在染的过程中发现有两连的两个顶点已经被染成相同的颜色，说明不是二分图。
-> 使用两种颜色，分别用1和-1来表示，初始时每个顶点用0表示未染色
-> 遍历每一个顶点，如果该顶点未被访问过，调用递归函数，如果返回false，说明不是二分图，直接返回false；
如果循环退出后没有返回false，则返回true。
-> 在递归函数中，如果当前顶点已经染色：如果该顶点的颜色和将要染的颜色相同，则返回true；否则返回false。
如果没被染色，则将当前顶点染色(-1*color: 将相邻点染成不同颜色)，然后再遍历与该顶点相连的所有的顶点，调用递归函数，如果返回false了，则当前递归函数的返回false，循环结束返回true.
time: O(V+E), space: O(V+E)
class Solution { int[] colors; public boolean isBipartite(int[][] graph) { int n = graph.length; colors = new int[n]; for(int i = 0; i < n; i++) { if(colors[i] == 0 && !dfs(graph, i, 1)) return false; } return true; } private boolean dfs(int[][] graph, int node, int color) { if(colors[node] != 0) return colors[node] == color ? true: false; else { colors[node] = color; for(int j : graph[node]) { if(!dfs(graph, j, -color)) return false; } } return true; } }