找到离给定两个节点最近的节点

/*
题目思路：
1. 使用广度优先搜索计算从node1和node2到其他节点的最短距离。
2. 使用Java Stream来处理节点的距离和找到最小的最大距离节点。
3. 如果没有满足条件的节点，返回-1。
*/

import java.util.*;
import java.util.stream.IntStream;

public class Solution {
    public int findClosestMeetingNode(int[] edges, int node1, int node2) {
        int n = edges.length;
        int[] dist1 = bfs(edges, node1);
        int[] dist2 = bfs(edges, node2);
        
        return IntStream.range(0, n)
                .filter(i -> dist1[i] < Integer.MAX_VALUE && dist2[i] < Integer.MAX_VALUE)
                .mapToObj(i -> new int[]{i, Math.max(dist1[i], dist2[i])})
                .min(Comparator.<int[]>comparingInt(a -> a[1]).thenComparingInt(a -> a[0]))
                .map(a -> a[0])
                .orElse(-1);
    }
    
    private int[] bfs(int[] edges, int start) {
        int n = edges.length;
        int[] dist = new int[n];
        Arrays.fill(dist, Integer.MAX_VALUE);
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(start);
        dist[start] = 0;
        
        while (!queue.isEmpty()) {
            int node = queue.poll();
            int neighbor = edges[node];
            if (neighbor != -1 && dist[neighbor] == Integer.MAX_VALUE) {
                dist[neighbor] = dist[node] + 1;
                queue.offer(neighbor);
            }
        }
        
        return dist;
    }
}