797. All Paths From Source to Target

最新推荐文章于 2020-12-30 23:28:42 发布

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本文介绍了一种使用深度优先搜索(DFS)算法解决图论中从节点0到节点N-1的所有可能路径的问题。通过一个具体示例，详细展示了如何在给定的有向无环图中找到所有路径，并保持路径内节点的正确顺序。

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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

图论知识，DFS求解所有路径，程序如下所示：

class Solution {
    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
        List<Integer> list = new ArrayList<>();
        List<List<Integer>> result = new ArrayList<>();
        dfs(graph, new boolean[graph.length], 0, list, result);
        return result;
    }
    
    public void dfs(int[][] graph, boolean[] visited, int cur, List<Integer> list, List<List<Integer>> result){
        if (cur > graph.length){
            return;
        }
        list.add(cur);
        if (cur == graph.length - 1){
            result.add(new ArrayList<Integer>(list));
            return;
        }
        visited[cur] = true;
        for (int v : graph[cur]){
            if (!visited[v]){
                dfs(graph, visited, v, list, result);
                list.remove(list.size() - 1);
            }
        }
        visited[cur] = false;
    }
}