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].

## Solutions

首先要弄懂输入是什么意思，输入表示的是下标从0 开始的结点的相邻结点有哪些，比如例子 1 中，第一个 list 中的[1,3]表示与结点 0 相邻的结点是 1 和 3，以后一次类推。

### 1. 染色法-递归

可以通过染色，如果两个相邻的结点被染成同一种颜色那么就不是二分图，否则便是。

class Solution:
def isBipartite(self, graph: List[List[int]]) -> bool:
colors = [0 for _ in range(len(graph))]
for i in range(len(colors)):
if colors[i] == 0 and not self.valid(graph, 1, i, colors):
return False
return True

def valid(self, graph, color, cur_pos, colors):
if colors[cur_pos] != 0:
return colors[cur_pos] == color
colors[cur_pos] = color
for i in graph[cur_pos]:
if not self.valid(graph, -1*color, i, colors):
return False
return True
# Runtime: 200 ms, faster than 95.60% of Python3 online submissions for Is Graph Bipartite?.
# Memory Usage: 14 MB, less than 9.09% of Python3 online submissions for Is Graph Bipartite?.