Grafik enthält einen Zyklus
/*
This is an implementation that determines
whether or not a directed graph contains a
cycle. A cycle is a number of vertices that
are connected in a closed chain.
Let V be the number of vertices and E the
number of edges.
Time complexity: O(V + E)
Space complexity: O(V)
*/
import java.util.List;
import java.util.ArrayList;
public class CycleInGraph {
private boolean cycle;
// WHITE: Vertex not processed yet
// GRAY: Vertex is being processed
// BLACK: Vertex already processed
private final int WHITE = 0, GRAY = 1, BLACK = 2;
private int[] colors;
private List<Integer[]> edges;
public CycleInGraph() {
cycle = false; // Cycle indicator
// Create list of Graph edges
edges = new ArrayList<>();
edges.add(new Integer[] { 1, 3 }); // Edges sourced at vertex 0
edges.add(new Integer[] { 2, 3, 4 }); // Edges sourced at 1
edges.add(new Integer[] { 0 }); // Edges sourced at vertex 2
edges.add(new Integer[] {}); // Edges sourced at vertex 3
edges.add(new Integer[] { 2, 5 }); // Edges sourced at vertex 4
edges.add(new Integer[] {}); // Edges sourced at vertex 5
// Colors array to assign a color per vertex
colors = new int[edges.size()];
}
public static void main(String[] args) {
CycleInGraph application = new CycleInGraph();
// There are multiple cycles in considered graph:
// 1) 0 -> 1 -> 2 -> 0
// 2) 0 -> 1 -> 4 -> 2 -> 0
// 3) 1 -> 2 -> 0 -> 1
System.out.println(application.cycleInGraph()); // true
}
public boolean cycleInGraph() {
for (int vertex = 0; vertex < edges.size(); vertex++) {
// Traverse graph using DFS algorithm
if (!cycle && colors[vertex] == WHITE) {
dfs(vertex);
}
}
return cycle;
}
private void dfs(int vertex) {
colors[vertex] = GRAY;
// Examine neighbors of current vertex
Integer[] neighbors = edges.get(vertex);
for (int neighbor : neighbors) {
// There is cycle if neighbor already
// processed in DFS traversal.
if (colors[neighbor] == GRAY) {
cycle = true;
return;
} else {
dfs(neighbor);
}
}
colors[vertex] = BLACK;
}
}
Wissam