- All Paths From Source to TargetGiven 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 v2—>3There 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<List<Integer>> result = new LinkedList<List<Integer>>(); public List<List<Integer>> allPathsSourceTarget(int[][] graph) { int index=0; for(int x:graph[index]){ List<Integer> record = new LinkedList<>(); record.add(0); record.add(x); depthFirst(graph, x, record); } return result; } public void depthFirst(int[][] graph, int index, List<Integer> 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<Integer> temp = new LinkedList<Integer>(); for(int i=0;i<size-1;i++){ temp.add(record.get(i)); } record = temp; } }}