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; } }
文章版权归作者所有,未经允许请勿转载,若此文章存在违规行为,您可以联系管理员删除。
转载请注明本文地址:https://www.ucloud.cn/yun/72294.html
Course Schedule I There are a total of n courses you have to take, labeled from 0 to n - 1.Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is e...
摘要:我们只要保证,对于第一次遇到的图节点,我们都会建立一个克隆节点,并在哈希表映射起来就行了。所以只要哈希表中有这个图节点,就说明我们之前已经将该图节点放入队列了,就不需要再处理了。 Clone Graph Clone an undirected graph. Each node in the graph contains a label and a list of its neighb...
Problem Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the ...
Problem Given a directed, acyclic graph of N nodes. Find all possible paths from node 0 to node N-1, and return them in any order. The graph is given as follows: the nodes are 0, 1, ..., graph.lengt...
阅读 1105·2023-04-25 17:28
阅读 3383·2021-10-14 09:43
阅读 3898·2021-10-09 10:02
阅读 1907·2019-08-30 14:04
阅读 3095·2019-08-30 13:09
阅读 3244·2019-08-30 12:53
阅读 2855·2019-08-29 17:11
阅读 1786·2019-08-29 16:58