785.Is Graph Bipartite?

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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].

判断一个图是否是二分图。这题给出了二分图的定义:即图上的顶点可以被分为互相独立的两簇,图上每条的两个顶点分别在这两簇中。做法可以是模拟这个定义,对图上每条边的顶点进行着色。如果图上每条边的顶点都能不冲突的着上两个色。则可以是二分图,解法可以是BFS也可以是DFS,下面给出DFS的解法:

class Solution(object):
    def isBipartite(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: bool
        """
        colorset = [-1] * len(graph)
#防止图不连通,所以要多次出发。
for i in xrange(len(graph)): if colorset[i] == -1 and not self.validcolor(graph, i, colorset, 0): return False return True def validcolor(self, graph, node, colorset, color): colorset[node] = color newcolor = 1 - color for n in graph[node]: if colorset[n] == -1 and not self.validcolor(graph, n, colorset, newcolor): return False elif colorset[n] != newcolor: return False return True

参考解法链接:https://leetcode.com/problems/is-graph-bipartite/discuss/115487/Java-Clean-DFS-solution-with-Explanation


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