［抄题］：
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.
[暴力解法]：
时间分析：
空间分析：
[优化后]：
时间分析：
空间分析：
［奇葩输出条件］：
［奇葩corner case］：
需要考虑非连通图的情况，此时无法染色 语言：
!validColor(graph, colors, 0, i)
［思维问题］：
如果valid函数中有染色, 那主函数就不用再染 直接用valid函数即可
［一句话思路］：
双色问题的本质是BFS
［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：
［画图］：
［一刷］：
- 理解下：下一个节点是用其index i表示的
［二刷］：
- “已经染色”用color[i] != -1来表示，同时检查一下有没有染对
［三刷］：
［四刷］：
［五刷］：
[五分钟肉眼debug的结果]：
［总结］：
如果valid函数中有染色, 那主函数就不用再染 直接用valid函数即可
［复杂度］：Time complexity: O(n^2) Space complexity: O(n^2)
［英文数据结构或算法，为什么不用别的数据结构或算法］：
初始化数组：
Arrays.fill(nums, -1);
［算法思想：递归/分治/贪心］：贪心
［关键模板化代码］：
public boolean validColor(int[][] graph, int[] colors, int color, int node) { //already colored，color again if (colors[node] != -1) return colors[node] == color; //color 1, then bfs colors[node] = color; for (int next : graph[node]) { if (!validColor(graph, colors, 1 - color, next)) return false; } return true; }
［其他解法］：
［Follow Up］：
［LC给出的题目变变变］：
[代码风格] ：
class Solution { public boolean isBipartite(int[][] graph) { //cc: null if (graph == null || graph.length == 0) return false; //ini: colors[] -1 int n = graph.length; int[] colors = new int[n]; Arrays.fill(colors, -1); //for loop: not connected graph for (int i = 0; i < n; i++) { if (colors[i] == -1 && !validColor(graph, colors, 0, i)) return false; } return true; } public boolean validColor(int[][] graph, int[] colors, int color, int node) { //already colored，color again if (colors[node] != -1) return colors[node] == color; //color 1, then bfs colors[node] = color; for (int next : graph[node]) { if (!validColor(graph, colors, 1 - color, next)) return false; } return true; } }