Hard
You are given a directed graph of n
nodes numbered from 0
to n - 1
, where each node has at most one outgoing edge.
The graph is represented with a given 0-indexed array edges
of size n
, indicating that there is a directed edge from node i
to node edges[i]
. If there is no outgoing edge from node i
, then edges[i] == -1
.
Return the length of the longest cycle in the graph. If no cycle exists, return -1
.
A cycle is a path that starts and ends at the same node.
Example 1:
Input: edges = [3,3,4,2,3]
Output: 3
Explanation: The longest cycle in the graph is the cycle: 2 -> 4 -> 3 -> 2.
The length of this cycle is 3, so 3 is returned.
Example 2:
Input: edges = [2,-1,3,1]
Output: -1
Explanation: There are no cycles in this graph.
Constraints:
n == edges.length
2 <= n <= 105
-1 <= edges[i] < n
edges[i] != i
public class Solution {
public int longestCycle(int[] edges) {
int n = edges.length;
boolean[] vis = new boolean[n];
boolean[] dfsvis = new boolean[n];
int[] path = new int[n];
int maxLength = -1;
for (int i = 0; i < n; i++) {
if (!vis[i]) {
path[i] = 1;
maxLength = Math.max(maxLength, dfs(i, 1, path, vis, dfsvis, edges));
}
}
return maxLength;
}
private int dfs(
int node, int pathLength, int[] path, boolean[] vis, boolean[] dfsvis, int[] edges) {
vis[node] = true;
dfsvis[node] = true;
int length = -1;
if (edges[node] != -1 && !vis[edges[node]]) {
path[edges[node]] = pathLength + 1;
length = dfs(edges[node], pathLength + 1, path, vis, dfsvis, edges);
} else if (edges[node] != -1 && dfsvis[edges[node]]) {
length = pathLength - path[edges[node]] + 1;
}
dfsvis[node] = false;
return length;
}
}