785. Is Graph Bipartite?

Leetcode Depth-first Search Breath-first Search Graph

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].

分析

判断二分图。详细分析和解答在Algorithms 4^th这本书中。

private boolean bipartitable;

public boolean isBipartite(int[][] graph) {
    bipartitable = true;
    boolean[] marked = new boolean[graph.length];
    boolean[] color  = new boolean[graph.length];
    for (int v = 0; v < graph.length; v++)
        if (bipartitable && !marked[v])
            dfs(graph, color, marked, v);
    return bipartitable;   
}

private void dfs(int[][] graph, boolean[] color, boolean[] marked, int v) {
    marked[v] = true;
    for (int w : graph[v]){
        if (!bipartitable) return;
        else if (!marked[w]) {
            color[w] = !color[v];
            dfs(graph, color, marked, w);
        }
        else if (color[w] == color[v]) bipartitable = false;
    }
}