Summary: Dectect cycle in directed graph: Detect cycle in a directed graph is using aDFS. Depth First Traversal can be used to detect cycle in a Graph. DFS for a connected graph produces a tree. There is a cycle in a graph only if there is aback edgepresent in the graph. A back edg...
力扣leetcode.cn/problems/distance-to-a-cycle-in-undirected-graph/ 无向图中仅包含一个环,想要直接定位图中的节点并不容易,但是有一类节点是很好区分的,就是度为1的节点,这些度为1的节点必然不在环中。如果将这些节点以及与之相连的边从图中移除,就可能得到另一批新的度为1的节点,那么只需重复这个过程...
int[][] edges = {{0, 1}, {1, 2}, {2, 0}}; boolean result = so.hasCycleDirectedGraph(3, edges); System.out.println(result); } } Detect Cycle in Undirected Graph 无向图找环 Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of n...
Count of Palindromic Substrings,最长子字符串的个数问题 Minimum Deletions in a String to make it a Palindrome,怎么删掉最少字符构成回文 Palindromic Partitioning,怎么分配字符,形成回文 5. Longest Common Substring,最长子字符串系列,13个题 Longest Common Substring,最长相同子串 Longest Common Subsequence,最长...
traverse(graph,0, visited, onPath);if(hasCycle) System.out.println("Has cycle!");returnres; }publicvoidtraverse(int[][] graph,intnode,boolean[] visited,boolean[] onPath){if(onPath[node]) hasCycle =true;if(visited[node])return;//在外循环修改visited和onPathvisited[node] =true; ...
/** * DAG's Single Source Shortest Path Algorithm in C++ * Based on DFS, don't let any circle exist in the graph * Time Cost : O(| 不高不富不帅的陈政_ 2018/05/18 9760 golang刷leetcode 经典(13) 最小高度树 queue 对于一个具有树特征的无向图,我们可选择任何一个节点作为根。图因此...
2192.All-Ancestors-of-a-Node-in-a-Directed-Acyclic-Graph (M) 2204.Distance-to-a-Cycle-in-Undirected-Graph (M) 2392.Build-a-Matrix-With-Conditions (M+) 2440.Create-Components-With-Same-Value (H-) 2603.Collect-Coins-in-a-Tree (H-) Dijkstra (BFS+PQ) 743.Network-Delay-Time (H-) 40...
1.leetcode 997: find-the-town-judge - 选定trust认为是一个从当前数字到n的线段,对于judge来说只有别的到它的线段,没有它对别人的线段 2.leetcode 1971:find-if-path-exists-in-graph - DFS遍历图,用ArrayList<ArrayList>存储图里面的指向关系,把source作为起点,看它的邻居里有没有destination ...
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} private boolean cycle(Graph graph, int[] visited){ if(visited[graph.val] == 1){ return true; } if(visited[graph.val] == 2){ return false; } visited[graph.val] = 1; for(Graph no : graph.neighbors){ if(cycle(no, visited)){ return true; } } visited[graph.val] = 2; retur...