描述
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 graph[i][j]).
Example 1:
Input: graph = [[1,2],[3],[3],[]]
Output: [[0,1,3],[0,2,3]]
Explanation: There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.
Example 2:
Input: graph = [[4,3,1],[3,2,4],[3],[4],[]]
Output: [[0,4],[0,3,4],[0,1,3,4],[0,1,2,3,4],[0,1,4]]
Note:
- n == graph.length
- 2 <= n <= 15
- 0 <= graph[i][j] < n
- graph[i][j] != i (i.e., there will be no self-loops).
- All the elements of graph[i] are unique.
- The input graph is guaranteed to be a DAG.
解析
根据题意，给定一个由 n 个标记为从 0 到 n - 1 的节点的有向无环图 （DAG），找到从节点 0 到节点 n - 1 的所有可能路径，并按任意顺序返回它们。
其实这道题就是考察 DFS ，直接使用 DFS 从 0 开始到最后一个标签的节点，把所有可能的路线走一遍然后把路线存入结果 resut 中就好了。
时间复杂度为 O(N*2^N) ，因为 DAG 中有 N 个节点，最坏情况下每一个节点都连接一个编号比它大的节点，此时路径数的量级为 O(2^N)，每条路径长度最长为 O(N)，空间复杂度为 O(N) ，主要是递归栈的开销。
解答
class Solution(object):
def allPathsSourceTarget(self, graph):
"""
:type graph: List[List[int]]
:rtype: List[List[int]]
"""
result = []
N = len(graph)
def dfs(node, L):
if node == N-1:
result.append(L)
return
for n in graph[node]:
dfs(n, L+[n])
dfs(0, [0])
return result
运行结果
Runtime 63 ms，Beats 99.25%
Memory 14.8 MB，Beats 45.48%
原题链接
https://leetcode.com/problems/all-paths-from-source-to-target/description/
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