797. 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.

思路

这题直接DFS,递归遍历就好了

• 时间复杂度 O(N)
• 空间复杂度 O(N)

代码

pub fn all_paths_source_target(graph: Vec<Vec<i32>>) -> Vec<Vec<i32>> { let mut paths = Vec::with_capacity(graph.len()); let mut path = Vec::with_capacity(graph.len()); path.push(0); find_N(0,&mut path,&mut paths,&graph); return paths; } pub fn find_N(poient:i32,mut path:&mut Vec<i32>,mut paths:&mut Vec<Vec<i32>>,graph:&Vec<Vec<i32>>){ if poient == (graph.len()-1) as i32 { paths.push(path.clone()); } else { for next in &graph[poient as usize]{ path.push(*next); find_N(*next,&mut path,&mut paths,& graph); path.pop(); } } }

• 执行用时: 12 ms
• 内存消耗: 2.4 MB

题型与相似题

题型

1.DFS
2.graph

相似题

• DFS
• 图
• 543. Diameter of Binary Tree

代码链接

all_paths_from_source_to_target