难度: Medium
原题连接
内容描述
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.
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].
思路 1 - 时间复杂度: O(V+E)- 空间复杂度: O(V)******
对于二分图中任意一点来说，它的neighbor都必定在另外一个集合，否则就不是二分图
BFS的思想
beats 100%
class Solution:
def isBipartite(self, graph):
"""
:type graph: List[List[int]]
:rtype: bool
"""
visited = set()
for i in range(len(graph)):
if i in visited or not graph[i]:
continue
cur_level = set([i])
while cur_level:
nxt_level = set()
for j in cur_level:
visited.add(j)
for n in graph[j]:
if n in cur_level:
return False
if n not in visited:
nxt_level.add(n)
cur_level = nxt_level
return True