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.

问0->n-1有几条路径

其实没什么,就是看输入有点不懂怎么建图,把图建立好,我们dfs就好了

class Solution {
public:
int Mp[][];
vector<vector<int>>ans;
vector<int>tmp;
void dfs(int n,int len){
if(len-==n){
ans.push_back(tmp);
return;
}
for(int i=;i<len;i++){
if(Mp[n][i]){
tmp.push_back(i);
dfs(i,len);
tmp.pop_back();
}
}
}
vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
int len=graph.size();
for(int i=;i<len;i++){
for(int j=;j<graph[i].size();j++){
Mp[i][graph[i][j]]=;
}
}
tmp.clear();
ans.clear();
tmp.push_back();
dfs(,len);
return ans; }
};

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