1557. Minimum Number of Vertices to Reach All Nodes

Medium

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.

Constraints:

• 2 <= n <= 10^5

• 1 <= edges.length <= min(10^5, n * (n - 1) / 2)

• edges[i].length == 2

• 0 <= fromi, toi < n

• All pairs (fromi, toi) are distinct.

解題

統計每個點的 indegree，indegree 等於零代表：如果不從該節點出發，則永遠走不到，因次加入答案。不等於零則是遲早會走到。

Runtime: 166 ms, faster than 100%

Memory Usage: 16.1 MB, less than 96.97%

func findSmallestSetOfVertices(n int, edges [][]int) []int {
    indegree := make([]int, n)
    for _, edge := range edges {
        indegree[edge[1]]++
    }

    ans := make([]int, 0)
    for i, n := range indegree {
        if n == 0 {
            ans = append(ans, i)
        }
    }

    return ans
}