1557. Minimum Number of Vertices to Reach All Nodes

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題目敘述

Given a directed acyclic graph, with n vertices numbered from 0 to n - 1, and an array edges where edges[i] = [fromi, toi] represents a directed edge from node fromi to node toi.

Find the smallest set of vertices from which all nodes in the graph are reachable. It’s guaranteed that a unique solution exists.

Notice that you can return the vertices in any order.

Example 1:

Input: n = 6, edges = [[0,1],[0,2],[2,5],[3,4],[4,2]] Output: [0,3] Explanation: It’s not possible to reach all the nodes from a single vertex. From 0 we can reach [0,1,2,5]. From 3 we can reach [3,4,2,5]. So we output [0,3].

Example 2:

Input: n = 5, edges = [[0,1],[2,1],[3,1],[1,4],[2,4]] Output: [0,2,3] Explanation: Notice that vertices 0, 3 and 2 are not reachable from any other node, so we must include them. Also any of these vertices can reach nodes 1 and 4.

解題思路

Solution

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public List<Integer> findSmallestSetOfVertices(int n, List<List<Integer>> edges) {
        Set<Integer> link = new HashSet<>();
        for (List<Integer> edge : edges) {
            link.add(edge.get(1));
        }
        
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (!link.contains(i)) {
                ans.add(i);
            }
        }
        
        return ans;
    }
}