N-1<\/code>, and return them in any order.<\/p>\nThe graph is given as follows:\u00a0 the nodes are 0, 1, …, graph.length – 1.\u00a0 graph[i] is a list of all nodes j for which the edge (i, j) exists.<\/p>\n
Example:\r\nInput: [[1,2], [3], [3], []] \r\nOutput: [[0,1,3],[0,2,3]] \r\nExplanation: The graph looks like this:\r\n0--->1\r\n| |\r\nv v\r\n2--->3\r\nThere are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.\r\n<\/pre>\nNote:<\/strong><\/p>\n\n- The number of nodes in the graph will be in the range\u00a0
[2, 15]<\/code>.<\/li>\n- You can print different paths in any order, but you should keep the order of nodes inside one path.<\/li>\n<\/ul>\n
Solution 1: DFS<\/strong><\/p>\nTime complexity: O(n!)<\/p>\n
Space complexity: O(n)<\/p>\n
\/\/ Author: Huahua\r\n\/\/ Running time: 85 ms\r\nclass Solution {\r\npublic:\r\n vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) { \r\n vector<vector<int>> ans;\r\n vector<int> path(1, 0); \r\n dfs(graph, 0, graph.size() - 1, path, ans);\r\n return ans;\r\n }\r\nprivate:\r\n void dfs(const vector<vector<int>>& graph, \r\n int cur, int dest, vector<int>& path, vector<vector<int>>& ans) {\r\n if (cur == dest) {\r\n ans.push_back(path);\r\n return;\r\n }\r\n \r\n for (int next : graph[cur]) { \r\n path.push_back(next);\r\n dfs(graph, next, dest, path, ans);\r\n path.pop_back();\r\n }\r\n }\r\n};<\/pre>\n“Cleaner” Version<\/p>\n
class Solution {\r\npublic:\r\n vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) { \r\n vector<vector<int>> ans;\r\n vector<int> path{0}; \r\n dfs(graph, path, ans);\r\n return ans;\r\n }\r\nprivate:\r\n void dfs(const vector<vector<int>>& graph, \r\n vector<int>& path, vector<vector<int>>& ans) {\r\n if (path.back() == graph.size() - 1) {\r\n ans.push_back(path);\r\n return;\r\n }\r\n \r\n for (int next : graph[path.back()]) {\r\n path.push_back(next);\r\n dfs(graph, path, ans);\r\n path.pop_back();\r\n }\r\n }\r\n};<\/pre>\n <\/p>\n","protected":false},"excerpt":{"rendered":"
\u9898\u76ee\u5927\u610f\uff1a\u7ed9\u4f60\u4e00\u4e2a\u65e0\u73af\u6709\u5411\u56fe\uff0c\u8fd4\u56de\u6240\u6709\u4ece\u8282\u70b90\u5230\u8282\u70b9n-1\u7684\u8def\u5f84\u3002 Problem: https:\/\/leetcode.com\/problems\/all-paths-from-source-to-target\/description Given a directed, acyclic graph of\u00a0N\u00a0nodes.\u00a0 Find all possible paths from node\u00a00\u00a0to node\u00a0N-1, and return them in any order. The graph is…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[76],"tags":[77,177],"class_list":["post-2057","post","type-post","status-publish","format-standard","hentry","category-graph","tag-graph","tag-medium","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/2057","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=2057"}],"version-history":[{"count":7,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/2057\/revisions"}],"predecessor-version":[{"id":2080,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/2057\/revisions\/2080"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=2057"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=2057"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=2057"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}