leetcode 1557. Minimum Number of Vertices to Reach All Nodes（python）

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描述

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

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.

Note:

2 <= n <= 10^5
1 <= edges.length <= min(10^5, n * (n - 1) / 2)
edges[i].length == 2
0 <= from_i, to_i < n
All pairs (from_i, to_i) are distinct.

解析

根据题意，就是给出了一个有向无环图，其中包含了 n 个节点，分别用索引 0 到 n-1 来表示节点，还有一个列表 edges ，其中每个元素都是一个子列表 [from_i, to_i] 表示从节点 from_i 到 to_i，题目要求我们找出最少的顶点个数，可以保证所有的节点都是可达的，这种节点集合是独一无二的，但是返回的结果顺序可以不同。

通过找规律可以发现，题目中要找的顶点都是只有出没有进的点，所以初始化 result 为所有节点的索引，然后遍历 edges 中的每个子列表 x ，只要 x[1] 在 result 说明其可达，所以从 result 中删掉该元素，遍历结束剩下的节点就是满足题意的顶点集合。

解答

class Solution(object):
    def findSmallestSetOfVertices(self, n, edges):
        """
        :type n: int
        :type edges: List[List[int]]
        :rtype: List[int]
        """
        result = set(range(n))
        for x,y in edges:
            if y in result:
                result.remove(y)
        return list(result)

运行结果

Runtime: 1036 ms, faster than 60.82% of Python online submissions for Minimum Number of Vertices to Reach All Nodes.
Memory Usage: 54.6 MB, less than 49.48% of Python online submissions for Minimum Number of Vertices to Reach All Nodes.

原题链接：leetcode.com/problems/mi…

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