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| 1 | +import java.util.ArrayList; |
| 2 | +import java.util.Arrays; |
| 3 | +import java.util.HashMap; |
| 4 | +import java.util.List; |
| 5 | +import java.util.Map; |
| 6 | +import java.util.PriorityQueue; |
| 7 | + |
| 8 | +public class ShortestPath { |
| 9 | + public int[] shortestPath(int n, int[][] edges, int start) { |
| 10 | + Map<Integer, List<int[]>> graph = new HashMap<>(); |
| 11 | + int[] distances = new int[n]; |
| 12 | + Arrays.fill(distances, Integer.MAX_VALUE); |
| 13 | + distances[start] = 0; |
| 14 | + // Represent the graph as an adjacency list. |
| 15 | + for (int[] edge : edges) { |
| 16 | + int u = edge[0]; |
| 17 | + int v = edge[1]; |
| 18 | + int w = edge[2]; |
| 19 | + |
| 20 | + graph.putIfAbsent(u, new ArrayList<>()); |
| 21 | + graph.putIfAbsent(v, new ArrayList<>()); |
| 22 | + graph.get(u).add(new int[]{v, w}); |
| 23 | + graph.get(v).add(new int[]{u, w}); |
| 24 | + } |
| 25 | + PriorityQueue<int[]> minHeap = new PriorityQueue<>((a, b) -> Integer.compare(a[0], b[0])); |
| 26 | + minHeap.offer(new int[]{0, start}); // (distance, node) |
| 27 | + // Use Dijkstra's algorithm to find the shortest path between the start node |
| 28 | + // and all other nodes. |
| 29 | + while (!minHeap.isEmpty()) { |
| 30 | + int[] curr = minHeap.poll(); |
| 31 | + int currDist = curr[0]; |
| 32 | + int currNode = curr[1]; |
| 33 | + // If the current distance to this node is greater than the recorded |
| 34 | + // distance, we've already found the shortest distance to this node. |
| 35 | + if (currDist > distances[currNode]) { |
| 36 | + continue; |
| 37 | + } |
| 38 | + // Update the distances of the neighboring nodes. |
| 39 | + for (int[] edge : graph.get(currNode)) { |
| 40 | + int neighbor = edge[0]; |
| 41 | + int weight = edge[1]; |
| 42 | + int neighborDist = currDist + weight; |
| 43 | + // Only update the distance if we find a shorter path to this |
| 44 | + // neighbor. |
| 45 | + if (neighborDist < distances[neighbor]) { |
| 46 | + distances[neighbor] = neighborDist; |
| 47 | + minHeap.offer(new int[]{neighborDist, neighbor}); |
| 48 | + } |
| 49 | + } |
| 50 | + } |
| 51 | + // Convert all infinity values to -1, representing unreachable nodes. |
| 52 | + for (int i = 0; i < n; i++) { |
| 53 | + if (distances[i] == Integer.MAX_VALUE) { |
| 54 | + distances[i] = -1; |
| 55 | + } |
| 56 | + } |
| 57 | + return distances; |
| 58 | + } |
| 59 | +} |
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