Problem
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].Solution DFS
class Solution { public boolean isBipartite(int[][] graph) { int[] colors = new int[graph.length]; for (int i = 0; i < graph.length; i++) { if (colors[i] == 0 && !dfs(graph, colors, i, 1)) return false; } return true; } private boolean dfs(int[][] graph, int[] colors, int node, int color) { if (colors[node] != 0) return colors[node] == color; else colors[node] = color; for (int neighbor: graph[node]) { if (!dfs(graph, colors, neighbor, -color)) return false; } return true; } }BFS
class Solution { public boolean isBipartite(int[][] graph) { int[] colors = new int[graph.length]; for (int i = 0; i < graph.length; i++) { if (graph[i].length != 0 && colors[i] == 0) { colors[i] = 1; Queuequeue = new LinkedList<>(); queue.offer(i); while (!queue.isEmpty()) { int node = queue.poll(); for (int neighbor: graph[node]) { if (colors[neighbor] == 0) { colors[neighbor] = -colors[node]; queue.offer(neighbor); } else { if (colors[neighbor] == colors[node]) return false; } } } } } return true; } }
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