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]]
Example 3:
Input: graph = [[1],[]] Output: [[0,1]]
Example 4:
Input: graph = [[1,2,3],[2],[3],[]] Output: [[0,1,2,3],[0,2,3],[0,3]]
Example 5:
Input: graph = [[1,3],[2],[3],[]] Output: [[0,1,2,3],[0,3]]
Constraints:
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.
所有可能的路径。
这个题是一道比较典型的把回溯backtracking应用到图的题目。既然是是找all possible paths,所以会想到类似DFS/backtracking的思路。面试遇到了不能做不出来啊。。。(# ̄~ ̄#)
这个题给的是array of array,题意不是很直观,我这里解释一下。比如题目中给的例子,意思是0可以连到1,2;1可以连接到3;2可以连接到3;3不能连到任何点(其实3也就是路径的终点了)。base case是当前遍历到graph里面的最后一个节点,则把当前list的结果加入结果集;其他部分则是跟其他经典的回溯类型的题没有区别。
时间O(n)
空间O(n)
Java实现
1 class Solution { 2 public List<List<Integer>> allPathsSourceTarget(int[][] graph) { 3 List<List<Integer>> res = new ArrayList<>(); 4 List<Integer> path = new ArrayList<>(); 5 path.add(0); 6 dfs(graph, 0, res, path); 7 return res; 8 } 9 10 private void dfs(int[][] graph, int index, List<List<Integer>> res, List<Integer> path) { 11 // base case 12 if (index == graph.length - 1) { 13 res.add(new ArrayList<>(path)); 14 return; 15 } 16 for (int next : graph[index]) { 17 path.add(next); 18 dfs(graph, next, res, path); 19 path.remove(path.size() - 1); 20 } 21 } 22 }