leetcode讲解--797. All Paths From Source to Target

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.

这一题考察的是图的深度优先遍历，要注意的点是：回溯的时候如何清除掉record中的数据，只保留到当前节点的数据，之后的数据需要重新填充。刚开始我是用的一个record，企图通过清空当前节点之后的数据，后来发现不对，record必须是多个，否则我给result添加的record就会全部相同。所以只能通过new和深拷贝来完成这个目标。

java代码

class Solution {
    List> result = new LinkedList>();
    public List> allPathsSourceTarget(int[][] graph) {
        int index=0;
        for(int x:graph[index]){
            List record = new LinkedList<>();
            record.add(0);
            record.add(x);
            depthFirst(graph, x, record);
        }
        return result;
    }
    
    public void depthFirst(int[][] graph, int index, List record){
        if(index==graph.length-1){
            result.add(record);
            return;
        }
        if(graph[index]==null){
            return;
        }
        for(int x:graph[index]){
            record.add(x);
            int size = record.size();
            depthFirst(graph, x, record);
            List temp = new LinkedList();
            for(int i=0;i