Leetcode 797. All Paths From Source to Target

标签：rom   描述   没有   ems   diff   print   out   int   elf   

题目链接：https://leetcode.com/problems/all-paths-from-source-to-target/description/

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.

看完题目描述，直觉就是DFS搜索解空间树。因为不是二维表格所以不好用DP，同时是无环图所以感觉比较像DFS。代码如下：

class Solution(object):
    def allPathsSourceTarget(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: List[List[int]]
        """
        res = []
        target = len(graph) - 1
        self.dfs([0], res, graph[0], graph, target)
        return res
        
    def dfs(self, curr_sol, res, curr_node, graph, target):
        if not curr_node:
            return
        for nxt in curr_node: 
            if nxt == target:
                res.append(curr_sol + [nxt])
            else:
                self.dfs(curr_sol+[nxt], res, graph[nxt], graph, target)

感觉是一道很标准的DFS，没有什么难点。

Leetcode 797. All Paths From Source to Target

标签：rom   描述   没有   ems   diff   print   out   int   elf   

原文地址：https://www.cnblogs.com/sylvanyang/p/9486400.html