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HierholzerAlgorithm.java
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140 lines (117 loc) · 4.33 KB
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package com.thealgorithms.graph;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.Stack;
/**
* Implementation of Hierholzer's algorithm to find an Eulerian Circuit in an undirected graph.
* <p>
* An Eulerian circuit is a trail in a graph that visits every edge exactly once,
* starting and ending at the same vertex. This algorithm finds such a circuit if one exists.
* </p>
* <p>
* This implementation is designed for an <strong>undirected graph</strong>. For a valid Eulerian
* circuit to exist, the graph must satisfy two conditions:
* <ol>
* <li>All vertices with a non-zero degree must be part of a single connected component.</li>
* <li>Every vertex must have an even degree (an even number of edges connected to it).</li>
* </ol>
* </p>
* <p>
* The algorithm runs in O(E + V) time, where E is the number of edges and V is the number of vertices.
* The graph is represented by a Map where keys are vertices and values are a LinkedList of adjacent vertices.
* </p>
*
* @see <a href="https://en.wikipedia.org/wiki/Eulerian_path#Hierholzer's_algorithm">Wikipedia: Hierholzer's algorithm</a>
*/
public final class HierholzerAlgorithm {
private final Map<Integer, LinkedList<Integer>> graph;
public HierholzerAlgorithm(Map<Integer, LinkedList<Integer>> graph) {
this.graph = (graph == null) ? new HashMap<>() : graph;
}
public boolean hasEulerianCircuit() {
if (graph.isEmpty()) {
return true;
}
for (List<Integer> neighbors : graph.values()) {
if (neighbors.size() % 2 != 0) {
return false;
}
}
return isCoherentlyConnected();
}
public List<Integer> findEulerianCircuit() {
if (!hasEulerianCircuit()) {
return Collections.emptyList();
}
Map<Integer, LinkedList<Integer>> tempGraph = new HashMap<>();
for (Map.Entry<Integer, LinkedList<Integer>> entry : graph.entrySet()) {
tempGraph.put(entry.getKey(), new LinkedList<>(entry.getValue()));
}
Stack<Integer> currentPath = new Stack<>();
LinkedList<Integer> circuit = new LinkedList<>();
int startVertex = -1;
for (Map.Entry<Integer, LinkedList<Integer>> entry : tempGraph.entrySet()) {
if (!entry.getValue().isEmpty()) {
startVertex = entry.getKey();
break;
}
}
if (startVertex == -1) {
if (graph.isEmpty()) {
return Collections.emptyList();
}
return Collections.singletonList(graph.keySet().iterator().next());
}
currentPath.push(startVertex);
while (!currentPath.isEmpty()) {
int currentVertex = currentPath.peek();
if (tempGraph.containsKey(currentVertex) && !tempGraph.get(currentVertex).isEmpty()) {
int nextVertex = tempGraph.get(currentVertex).pollFirst();
tempGraph.get(nextVertex).remove(Integer.valueOf(currentVertex));
currentPath.push(nextVertex);
} else {
circuit.addFirst(currentVertex);
currentPath.pop();
}
}
return circuit;
}
private boolean isCoherentlyConnected() {
if (graph.isEmpty()) {
return true;
}
Set<Integer> visited = new HashSet<>();
int startNode = -1;
for (Map.Entry<Integer, LinkedList<Integer>> entry : graph.entrySet()) {
if (!entry.getValue().isEmpty()) {
startNode = entry.getKey();
break;
}
}
if (startNode == -1) {
return true;
}
dfs(startNode, visited);
for (Map.Entry<Integer, LinkedList<Integer>> entry : graph.entrySet()) {
if (!entry.getValue().isEmpty() && !visited.contains(entry.getKey())) {
return false;
}
}
return true;
}
private void dfs(int u, Set<Integer> visited) {
visited.add(u);
if (graph.containsKey(u)) {
for (int v : graph.get(u)) {
if (!visited.contains(v)) {
dfs(v, visited);
}
}
}
}
}