797 All Paths From Source to Target

1. Question

Given a directed, acyclic graph ofNnodes. Find all possible paths from node0to nodeN-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.
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Example:
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Input: [[1,2], [3], [3], []]
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Output: [[0,1,3],[0,2,3]]
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Explanation:
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The graph looks like this:
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0--->1
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| |
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v v
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2--->3
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There are two paths:
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0 -> 1 ->3 and 0 -> 2 ->3.
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2. Implementation

(1) Backtracking
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class Solution {
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public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
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List<List<Integer>> res = new ArrayList<>();
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if (graph == null || graph.length == 0) {
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return res;
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}
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int m = graph.length, n = graph[0].length;
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List<Integer> path = new ArrayList<>();
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path.add(0);
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getPaths(0, graph, path, res);
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return res;
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}
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public void getPaths(int curNode, int[][] graph, List<Integer> path, List<List<Integer>> res) {
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if (curNode == graph.length - 1) {
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res.add(new ArrayList<>(path));
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return;
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}
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for (int nextNode : graph[curNode]) {
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path.add(nextNode);
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getPaths(nextNode, graph, path, res);
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path.remove(path.size() - 1);
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}
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}
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}
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3. Time & Space Complexity

Backtracking: 时间复杂度O(N * 2^N ), 总共有2^N个可能的路径, 而每个路径path.remove()需要O(N)的时间, 空间复杂度O(N * 2^N),总共有2 ^ N的路径,每个路径的递归深度最多为N