判断二分图

Problem

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.

Idea

A graph is bipartite exactly when we can color every node with one of two colors so that the two ends of every edge always have different colors.

Use BFS to do the coloring. Start from an uncolored node, give it color 0, and give each uncolored neighbor the opposite color of the current node. If a neighbor already has the same color as the current node, the graph is not bipartite. Since the graph may not be connected, start a new BFS from every node that is still uncolored. DFS works the same way.

Time complexity: \(O(n + m)\), Space complexity: \(O(n)\), where \(n\) is the number of nodes and \(m\) is the number of edges

Pseudo Code

color = array of size n, filled with -1;      // -1 means uncolored
for (start = 0; start < n; start++) {
    if (color[start] != -1) {
        continue;                              // already colored by an earlier BFS
    }
    color[start] = 0;
    queue = empty queue;
    queue.push(start);
    while (!queue.isEmpty()) {
        u = queue.pop();
        for (v in graph[u]) {
            if (color[v] == -1) {
                color[v] = 1 - color[u];       // opposite color
                queue.push(v);
            }
            else if (color[v] == color[u]) {
                return false;
            }
        }
    }
}
return true;