Link: https://leetcode.cn/problems/is-graph-bipartite/

Question

difficulty: mid
adj diff: 5

There is an undirected graph with n nodes, where each node is numbered between 0 and n - 1. You are given a 2D array graph, where graph[u] is an array of nodes that node u is adjacent to. More formally, for each v in graph[u], there is an undirected edge between node u and node v. The graph has the following properties:

	There are no self-edges (graph[u] does not contain u).
	There are no parallel edges (graph[u] does not contain duplicate values).
	If v is in graph[u], then u is in graph[v] (the graph is undirected).
	The graph may not be connected, meaning there may be two nodes u and v such that there is no path between them.

A graph is bipartite if the nodes can be partitioned into two independent sets A and B such that every edge in the graph connects a node in set A and a node in set B.

Return true if and only if it is bipartite.

Example 1:

Input: graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
Output: false
Explanation: There is no way to partition the nodes into two independent sets such that every edge connects a node in one and a node in the other.

Example 2:

Input: graph = [[1,3],[0,2],[1,3],[0,2]]
Output: true
Explanation: We can partition the nodes into two sets: {0, 2} and {1, 3}.

Constraints:

	graph.length == n
	1 <= n <= 100
	0 <= graph[u].length < n
	0 <= graph[u][i] <= n - 1
	graph[u] does not contain u.
	All the values of graph[u] are unique.
	If graph[u] contains v, then graph[v] contains u.

这道题的 bfs 搜索，看似不难，但是难在 graph 会有好几个孤岛。

如何处理孤岛，保证全都 visit 一遍，这个很难。

最后是 3 层循环，其中第二层循环之前才创建 queue。这个方法第一次碰到。写起来真费劲。

Code

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public boolean isBipartite(int[][] graph) {
    int len = graph.length;
    int[] visited = new int[len];

    for (int i = 0; i < len; i++) {
        if (visited[i] != 0) {
            continue;
        }
        List<Integer> list = new LinkedList<Integer>();
        list.add(i);
        visited[i] = 1;
        while (!list.isEmpty()) {
            int visitIndex = list.remove(0);
            int nextValue = 3 - visited[visitIndex];
            for (int visitNext: graph[visitIndex]) {
                // visitNext is adjacent to visitIndex, thus value should = nextValue
                if (visited[visitNext] == nextValue) {
                    continue;
                } else if (visited[visitNext] == 0) {
                    visited[visitNext] = nextValue;
                    list.add(visitNext);
                } else if (visited[visitNext] != nextValue) {
                    return false;
                }
            }
        }
    }
    return true;
}