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.

 

这道题给了我们一个无回路有向图,包含N个结点,然后让我们找出所有可能的从结点0到结点N-1的路径。这个图的数据是通过一个类似邻接链表的二维数组给的,最开始的时候博主没看懂输入数据的意思,其实很简单,我们来看例子中的input,[[1,2], [3], [3], []],这是一个二维数组,最外层的数组里面有四个小数组,每个小数组其实就是和当前结点相通的邻结点,由于是有向图,所以只能是当前结点到邻结点,反过来不一定行。那么结点0的邻结点就是结点1和2,结点1的邻结点就是结点3,结点2的邻结点也是3,结点3没有邻结点。那么其实这道题的本质就是遍历邻接链表,由于要列出所有路径情况,那么递归就是不二之选了。我们用cur来表示当前遍历到的结点,初始化为0,然后在递归函数中,先将其加入路径path,如果cur等于N-1了,那么说明到达结点N-1了,将path加入结果res。否则我们再遍历cur的邻接结点,调用递归函数即可,参见代码如下:

 

解法一:

class Solution {
public:
    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        vector<vector<int>> res;
        helper(graph, 0, {}, res);
        return res;
    }
    void helper(vector<vector<int>>& graph, int cur, vector<int> path, vector<vector<int>>& res) {
        path.push_back(cur);
        if (cur == graph.size() - 1) res.push_back(path);
        else for (int neigh : graph[cur]) helper(graph, neigh, path, res);
    }
};

 

下面这种解法也是递归,不过写法稍有不同,递归函数直接返回结果,这样参数就少了许多,但是思路还是一样的,如果cur等于N-1了,直接将cur先装入数组,再装入结果res中返回。否则就遍历cur的邻接结点,对于每个邻接结点,先调用递归函数,然后遍历其返回的结果,对于每个遍历到的path,将cur加到数组首位置,然后将path加入结果res中即可,这有点像是回溯的思路,路径是从后往前组成的,参见代码如下:

 

解法二:

class Solution {
public:
    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        return helper(graph, 0);
    }
    vector<vector<int>> helper(vector<vector<int>>& graph, int cur) {
        if (cur == graph.size() - 1) {
            return {{graph.size() - 1}};
        }
        vector<vector<int>> res;
        for (int neigh : graph[cur]) {
            for (auto path : helper(graph, neigh)) {
                path.insert(path.begin(), cur);
                res.push_back(path);
            }
        }
        return res;
    }
};

 

类似题目:

https://leetcode.com/problems/all-paths-from-source-to-target/solution/

https://leetcode.com/problems/all-paths-from-source-to-target/discuss/121135/6-lines-C++-dfs

https://leetcode.com/problems/all-paths-from-source-to-target/discuss/118691/Easy-and-Concise-DFS-Solution-C++-2-line-Python

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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