797. All Paths From Source to Target

Given a directed, acyclic graph of N nodes.  Find all possible paths from node 0 to node N-1, and return them in any order.

The graph is given as follows:  the nodes are 0, 1, ..., graph.length - 1.  graph[i] is a list of all nodes j for which the edge (i, j) exists.

Example:
Input: [[1,2], [3], [3], []] 
Output: [[0,1,3],[0,2,3]] 
Explanation: The graph looks like this:
0--->1
|    |
v    v
2--->3
There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.

Note:

• The number of nodes in the graph will be in the range [2, 15].
• You can print different paths in any order, but you should keep the order of nodes inside one path.

---

 有向无环图找到从起点到终点的所有路径，用DFS解决。

DFS：把source点加入path，查看是不是target点，if yes，把path加入result；otherwise，查看该点的邻居，recursively call DFS on each neighbor.

 1 class Solution {
 2 public:
 3     vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
 4         vector<vector<int>>result;
 5         if(graph.size()==0){
 6             return result;
 7         }
 8         vector<int> path;
 9         dfs(result,path,0,graph.size()-1,graph);
10         return result;
11     }
12 private:
13     void dfs(vector<vector<int>>&result, vector<int>&path,int source, int target, vector<vector<int>>& graph){
14         path.push_back(source);
15         if(source==target){
16             result.push_back(path);
17         }else{
18             for(int next: graph[source]){
19                 dfs(result,path,next,target,graph);
20             }
21         }
22         path.pop_back();
23     }
24 };

 

转载于:https://www.cnblogs.com/ruisha/p/9222277.html