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]).
Solution
求一个点到终点的每条路径。对于边的存储我们依然是用 /(vector/)，为了遍历每一种情况，在 /(DFS/) 的时候，记得 /(pop/_back/)
点击查看代码
class Solution {
private:
vector<vector<int>> ans;
void dfs(int from, int tgt, vector<vector<int>> graph, vector<int>& path, vector<vector<int>>& ans){
path.push_back(from);
if(from==tgt){
ans.push_back(path); return;
}
for(auto to: graph[from]){
dfs(to, tgt, graph, path, ans);
path.pop_back();
}
}
public:
vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
int sz = graph.size();
vector<int> path;
dfs(0, sz-1, graph, path, ans);
return ans;
}
};
原创文章，作者：ItWorker，如若转载，请注明出处：https://blog.ytso.com/279302.html