leetcode 797. All Paths From Source to Target（source到target的所有路径）

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原创最新推荐文章于 2026-03-06 08:23:23 发布 275 阅读 AI 写同款· GEO 优化 ›

专有钉钉小程序(浙政钉)开发搭建流程总览最近公司有一个项目要做浙政钉，让我来负责做。靠，我一脸懵，开发了两年多不知道浙政钉是啥，感觉我眼界还是低了。然后就去百度、同事问了一圈才知道就是个钉钉小程序嘛，然后发布到浙政钉上了。就搭建项目跑起来我都肥了很多事，中间走了很多坑。正好领导让我写《钉钉小程序前端开发流程》，这想着也写个博客吧。下面就是整个流程及我所遇到的坑！阅读详情

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:
在这里插入图片描述
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.

Constraints:
n == graph.length
2 <= n <= 15
0 <= graph[i][j] < n
graph[i][j] != i (i.e., there will be no self-loops).
The input graph is guaranteed to be a DAG.

有向无环图中有n个node，找出0到n-1节点的所有路径

思路：
从0出发DFS

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

由于不用遍历所有节点，而且是没有环的有向图，可以不用visited数组记访问过的位置

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