{ "data": { "question": { "questionId": "1431", "questionFrontendId": "2192", "categoryTitle": "Algorithms", "boundTopicId": 1308876, "title": "All Ancestors of a Node in a Directed Acyclic Graph", "titleSlug": "all-ancestors-of-a-node-in-a-directed-acyclic-graph", "content": "
You are given a positive integer n
representing the number of nodes of a Directed Acyclic Graph (DAG). The nodes are numbered from 0
to n - 1
(inclusive).
You are also given a 2D integer array edges
, where edges[i] = [fromi, toi]
denotes that there is a unidirectional edge from fromi
to toi
in the graph.
Return a list answer
, where answer[i]
is the list of ancestors of the ith
node, sorted in ascending order.
A node u
is an ancestor of another node v
if u
can reach v
via a set of edges.
\n
Example 1:
\n\n\nInput: n = 8, edgeList = [[0,3],[0,4],[1,3],[2,4],[2,7],[3,5],[3,6],[3,7],[4,6]]\nOutput: [[],[],[],[0,1],[0,2],[0,1,3],[0,1,2,3,4],[0,1,2,3]]\nExplanation:\nThe above diagram represents the input graph.\n- Nodes 0, 1, and 2 do not have any ancestors.\n- Node 3 has two ancestors 0 and 1.\n- Node 4 has two ancestors 0 and 2.\n- Node 5 has three ancestors 0, 1, and 3.\n- Node 6 has five ancestors 0, 1, 2, 3, and 4.\n- Node 7 has four ancestors 0, 1, 2, and 3.\n\n\n
Example 2:
\n\n\nInput: n = 5, edgeList = [[0,1],[0,2],[0,3],[0,4],[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]\nOutput: [[],[0],[0,1],[0,1,2],[0,1,2,3]]\nExplanation:\nThe above diagram represents the input graph.\n- Node 0 does not have any ancestor.\n- Node 1 has one ancestor 0.\n- Node 2 has two ancestors 0 and 1.\n- Node 3 has three ancestors 0, 1, and 2.\n- Node 4 has four ancestors 0, 1, 2, and 3.\n\n\n
\n
Constraints:
\n\n1 <= n <= 1000
0 <= edges.length <= min(2000, n * (n - 1) / 2)
edges[i].length == 2
0 <= fromi, toi <= n - 1
fromi != toi
给你一个正整数 n
,它表示一个 有向无环图 中节点的数目,节点编号为 0
到 n - 1
(包括两者)。
给你一个二维整数数组 edges
,其中 edges[i] = [fromi, toi]
表示图中一条从 fromi
到 toi
的单向边。
请你返回一个数组 answer
,其中 answer[i]
是第 i
个节点的所有 祖先 ,这些祖先节点 升序 排序。
如果 u
通过一系列边,能够到达 v
,那么我们称节点 u
是节点 v
的 祖先 节点。
\n\n
示例 1:
\n\n\n\n输入:n = 8, edgeList = [[0,3],[0,4],[1,3],[2,4],[2,7],[3,5],[3,6],[3,7],[4,6]]\n输出:[[],[],[],[0,1],[0,2],[0,1,3],[0,1,2,3,4],[0,1,2,3]]\n解释:\n上图为输入所对应的图。\n- 节点 0 ,1 和 2 没有任何祖先。\n- 节点 3 有 2 个祖先 0 和 1 。\n- 节点 4 有 2 个祖先 0 和 2 。\n- 节点 5 有 3 个祖先 0 ,1 和 3 。\n- 节点 6 有 5 个祖先 0 ,1 ,2 ,3 和 4 。\n- 节点 7 有 4 个祖先 0 ,1 ,2 和 3 。\n\n\n
示例 2:
\n\n\n\n输入:n = 5, edgeList = [[0,1],[0,2],[0,3],[0,4],[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]\n输出:[[],[0],[0,1],[0,1,2],[0,1,2,3]]\n解释:\n上图为输入所对应的图。\n- 节点 0 没有任何祖先。\n- 节点 1 有 1 个祖先 0 。\n- 节点 2 有 2 个祖先 0 和 1 。\n- 节点 3 有 3 个祖先 0 ,1 和 2 。\n- 节点 4 有 4 个祖先 0 ,1 ,2 和 3 。\n\n\n
\n\n
提示:
\n\n1 <= n <= 1000
0 <= edges.length <= min(2000, n * (n - 1) / 2)
edges[i].length == 2
0 <= fromi, toi <= n - 1
fromi != toi
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