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[12/30] 797. All Paths From Source to Target

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작성일 2022.12.30 19:30

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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<li style="border: 0px solid; box-sizing: border-box; --tw-border-spacing-x:0; --tw-border-spacing-y:0; --tw-translate-x:0; --tw-translate-y:0; --tw-rotate:0; --tw-skew-x:0; --tw-skew-y:0; --tw-scale-x:1; --tw-scale-y:1; --tw-pan-x: ; --tw-pan-y: ; --tw-pinch-zoom: ; --tw-scroll-snap-strictness:proximity; --tw-ordinal: ; --tw-slashed-zero: ; --tw-numeric-figure: ; --tw-numeric-spacing: ; --tw-numeric-fraction: ; --tw-ring-inset: ; --tw-ring-offset-width:0px; --tw-ring-offset-color:#fff; --tw-ring-color:rgba(59,130,246,0.5); --tw-ring-offset-shadow:0 0 #0000; --tw-ring-shadow:0 0 #0000; --tw-shadow:0 0 #0000; --tw-shadow-colored:0 0 #0000; --tw-blur: ; --tw-brightness: ; --tw-contrast: ; --tw-grayscale: ; --tw-hue-rotate: ; --tw-invert: ; --tw-saturate: ; --tw-sepia: ; --tw-drop-shadow: ; --tw-backdrop-blur: ; --tw-backdrop-brightness: ; --tw-backdrop-contrast: ; --tw-backdrop-grayscale: ; -

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링크

https://leetcode.com/problems/all-paths-from-source-to-target/description/ 827 회 연결

익명
작성일 2022.12.30 19:30

class Solution:
    def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
        res = []
        N = len(graph)-1

        path = [0]
        def dfs(curr):
            if curr == N:
                res.append(path.copy())
                return

            for nxt in graph[curr]:
                path.append(nxt)
                dfs(nxt)
                path.pop()
        
        dfs(0)
        return res