Given a directed acyclic graph (DAG) of
n nodes labeled from 0 to n – 1, find all possible paths from node
0 to node
n - 1, and return them in any order.
The graph is given as follows:
graph[i] is a list of all nodes you can visit from node
i (i.e., there is a directed edge from node
i to node
Input: graph = [[1,2],,,] Output: [[0,1,3],[0,2,3]] Explanation: There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.
Input: graph = [[4,3,1],[3,2,4],,,] Output: [[0,4],[0,3,4],[0,1,3,4],[0,1,2,3,4],[0,1,4]]
Input: graph = [,] Output: [[0,1]]
Input: graph = [[1,2,3],,,] Output: [[0,1,2,3],[0,2,3],[0,3]]
Input: graph = [[1,3],,,] Output: [[0,1,2,3],[0,3]]
n == graph.length
2 <= n <= 15
0 <= graph[i][j] < n
graph[i][j] != i(i.e., there will be no self-loops).
- The input graph is guaranteed to be a DAG.
Build an adjacency list representing the relationship between nodes and edges. Then conduct a depth-first search to find all the paths from source to target.
class Solution: def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]: results =  adjacency_list =  for i in graph: adjacency_list.append() for i in range(len(graph)): edges = graph[i] for edge in edges: adjacency_list[i].append(edge) self.helper(results, adjacency_list, 0, len(adjacency_list) - 1, ) return results def helper(self, results, adjacency_list, current, target, path): if current == target: results.append(list(path)) return for node in adjacency_list[current]: path.append(node) self.helper(results, adjacency_list, node, target, path) path.pop()
- Time Complexity: O(V + E). V is the number of vertices, E is the number of edges.
- Space Complexity: O(V + E). V is the number of vertices, E is the number of edges.