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785.Is Graph Bipartite?

发布时间:2020-12-14 04:20:52 所属栏目:大数据 来源:网络整理
导读: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 an

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],[1,2]]
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: [[1,2,1,2]]
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,color): colorset[node] = color newcolor = 1 - color for n in graph[node]: if colorset[n] == -1 and not self.validcolor(graph,n,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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