797. All Paths From Source to Target

Medium (Trung bình) C++ Python 🔗 Xem trên LeetCode

📋 Đề Bài

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]]

 

Constraints:

  • 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.

🧠 Thuật Toán & Kỹ Thuật

DFS (Tìm kiếm theo chiều sâu)Graph (Đồ thị)Bit Manipulation (Thao tác bit)Backtracking (Quay lui)
⏱️ Thời gian O(V+E)
💾 Không gian O(V)

💻 Lời Giải

C++ 0797-all-paths-from-source-to-target.cpp
class Solution {
private:
    int n;
    vector<vector<int>> ans, adj;
    vector<int> tmp;
    
public:
    void backTracking(vector<vector<int>>& graph, int u) {
        if (u == n - 1) {
            ans.push_back(tmp);
            return;
        }
        for (int v : graph[u]) {
            tmp.push_back(v);
            backTracking(graph, v);
            tmp.pop_back();
        }
    }
    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        this->n = graph.size();
        adj.resize(n);
        tmp.push_back(0);
        backTracking(graph, 0);
        return ans;
    }
};
Python 0797-all-paths-from-source-to-target.py
class Solution:
    def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
        ans = []
        n = len(graph)

        def dfs(u: int, path: List[int]) -> None:
            if u == n - 1:
                ans.append(path)
                return

            for v in graph[u]:
                dfs(v, path + [v])

        dfs(0, [0])

        return ans