节点间通路

/*
 * 题目思路：
 * 使用Java Stream进行图的DFS实现。通过流式处理减少显式循环的使用。
 * 通过递归的方式来检查从起始节点到目标节点是否存在路径。
 */
import java.util.*;
import java.util.stream.*;

public class SolutionStream {
    public boolean hasPath(int n, int[][] graph, int start, int target) {
        // 使用邻接表表示图
        Map<Integer, List<Integer>> adjList = Arrays.stream(graph)
                .collect(Collectors.groupingBy(
                        edge -> edge[0],
                        Collectors.mapping(edge -> edge[1], Collectors.toList())));

        // 标记节点是否访问过
        boolean[] visited = new boolean[n];
        return dfs(adjList, visited, start, target);
    }

    private boolean dfs(Map<Integer, List<Integer>> adjList, boolean[] visited, int current, int target) {
        if (current == target) {
            return true;
        }
        visited[current] = true;
        return adjList.getOrDefault(current, Collections.emptyList()).stream()
                .filter(neighbor -> !visited[neighbor])
                .anyMatch(neighbor -> dfs(adjList, visited, neighbor, target));
    }
}