[LeetCode] 797. All Paths From Source to Target

xiaochao 發(fā)布于2019-08-16 13:49 / 867人閱讀

Problem

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.

Solution - DFS

class Solution {
    public List> allPathsSourceTarget(int[][] graph) {
        List> res = new ArrayList<>();
        List temp = new ArrayList<>();
        int firstNode = 0;
        temp.add(firstNode);
        dfs(graph, firstNode, temp, res);
        return res;
    }
    
    private void dfs(int[][] graph, int node, List temp, List> res) {
        if (node == graph.length-1) {
            res.add(new ArrayList<>(temp));
            return;
        }
        for (int neighbor: graph[node]) {
            temp.add(neighbor);
            dfs(graph, neighbor, temp, res);
            temp.remove(temp.size()-1);
        }
    }
}

Solution - BFS

class Solution {
    public List> allPathsSourceTarget(int[][] graph) {
        int n = graph.length;
        List> res = new ArrayList<>();
        Deque> queue = new ArrayDeque<>();
        queue.offer(Arrays.asList(0));
        
        while (!queue.isEmpty()) {
            List cur = queue.poll();
            int size = cur.size();
            if (cur.get(size-1) == n-1) {
                res.add(cur);
                continue;
            }
            
            for (int node: graph[cur.get(size-1)]) {
                List next = new ArrayList<>(cur);
                next.add(node);
                queue.offer(next);
            }
        }
        
        return res;
    }
}