2018年9月23日
#785 Is Graph Bipartite?
问题描述：
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.
样例:
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:
graphwill have length in range[1, 100].graph[i]will contain integers in range[0, graph.length - 1].graph[i]will not containior duplicate values.- The graph is undirected: if any element
jis ingraph[i], theniwill be ingraph[j].
问题分析：
本题难度为Medium！已给出的函数定义为：
class Solution:def isBipartite(self, graph):""":type graph: List[List[int]]:rtype: bool"""其中graph为一个二维数组，返回值为布尔类型值；
本题是判断一个无向图是否为二分图，判断二分图方法有很多，这里采用算法实现上比较简单的染色法。
算法如下：
用colornodes列表记录结点的染色情况（-1/1），用visited列表记录结点的访问情况（0/1）
通过广度优先搜索bfs遍历每一个结点：
建立bfslist列表；
while 有结点未被访问：
若bfslist为空，则选则一个未访问的结点，染色，标记该结点已访问，并添加到bfslist后；
若bfslist不为空，则遍历该结点的子结点：
若该子结点未被访问，则染色（与父结点颜色不同），标记为已访问，并添加到bsflist后；
遍历graph表，若存在相邻结点颜色相同，则不是二分图，否则，是二分图；
代码实现：
#coding=utf-8
class Solution:def isBipartite(self, graph):""":type graph: List[List[int]]:rtype: bool"""visited=[0 for i in range(len(graph))] #访问过的结点标记为1 colornodes=[0 for i in range(len(graph))] #store the color of each node,-1/1curnode=0 #当前结点color=-1 #初始颜色为-1bfslist=[] #bfs队列while sum(visited)!=len(graph):if len(bfslist)==0: #bfs队列为空i=visited.index(0) #未访问过的结点visited[i]=1 #标记已访问colornodes[i]=color #染色bfslist.append(i) #加到bfs队列中else:curnode=bfslist.pop(0) #pop第一个元素color = colornodes[curnode]*-1# 切换染色颜色for i in graph[curnode]:if visited[i]==0: #处理未访问过的结点visited[i]=1 #标记已访问colornodes[i]=color #染色bfslist.append(i) #加到bfs队列中for i in range(len(graph)):for j in range(len(graph[i])):if colornodes[i]==colornodes[graph[i][j]]:return Falsereturn True