75th LeetCode Weekly Contest All Paths From Source to Target
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[30][30]; vector<vector<int>>ans; vector<int>tmp; void dfs(int n,int len){ if(len-1==n){ ans.push_back(tmp); return; } for(int i=0;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=0;i<len;i++){ for(int j=0;j<graph[i].size();j++){ Mp[i][graph[i][j]]=1; } } tmp.clear(); ans.clear(); tmp.push_back(0); dfs(0,len); return ans; } };
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 从 HTTP 原因短语缺失研究 HTTP/2 和 HTTP/3 的设计差异
· AI与.NET技术实操系列:向量存储与相似性搜索在 .NET 中的实现
· 基于Microsoft.Extensions.AI核心库实现RAG应用
· Linux系列:如何用heaptrack跟踪.NET程序的非托管内存泄露
· 开发者必知的日志记录最佳实践
· TypeScript + Deepseek 打造卜卦网站:技术与玄学的结合
· Manus的开源复刻OpenManus初探
· AI 智能体引爆开源社区「GitHub 热点速览」
· 从HTTP原因短语缺失研究HTTP/2和HTTP/3的设计差异
· 三行代码完成国际化适配,妙~啊~