<p>You are given a <strong>directed</strong> graph of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code>, where each node has <strong>at most one</strong> outgoing edge.</p>

<p>The graph is represented with a given <strong>0-indexed</strong> array <code>edges</code> of size <code>n</code>, indicating that there is a directed edge from node <code>i</code> to node <code>edges[i]</code>. If there is no outgoing edge from node <code>i</code>, then <code>edges[i] == -1</code>.</p>

<p>Return <em>the length of the <strong>longest</strong> cycle in the graph</em>. If no cycle exists, return <code>-1</code>.</p>

<p>A cycle is a path that starts and ends at the <strong>same</strong> node.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/08/graph4drawio-5.png" style="width: 335px; height: 191px;" />
<pre>
<strong>Input:</strong> edges = [3,3,4,2,3]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest cycle in the graph is the cycle: 2 -&gt; 4 -&gt; 3 -&gt; 2.
The length of this cycle is 3, so 3 is returned.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/07/graph4drawio-1.png" style="width: 171px; height: 161px;" />
<pre>
<strong>Input:</strong> edges = [2,-1,3,1]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There are no cycles in this graph.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == edges.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>-1 &lt;= edges[i] &lt; n</code></li>
	<li><code>edges[i] != i</code></li>
</ul>

class Solution {
public:
    vector<int> vis, nexts;
    int dfs(int node){
        if(vis[node]) return 0;
        vis[node] = 1;
        return 1 + dfs(nexts[node]);
    }

    int longestCycle(vector<int>& edges) {
        int n = edges.size();
        nexts = edges;
        vector<int> inDeg(n, 0);
        vis.resize(n, 0);
        for(int i=0; i<n; ++i)
            if(edges[i] != -1)  ++inDeg[edges[i]];

        queue<int> q;
        for(int i=0; i<n; ++i)
            if(inDeg[i] == 0)  q.push(i), vis[i] = 1;

        while(!q.empty()){
            int f = q.front();
            q.pop();
            int next = edges[f];
            if(next == -1) continue;
            if(--inDeg[next] == 0) q.push(next), vis[next] = 1;
        }

        int ans = 0;
        for(int i=0; i<n; ++i)
            if(!vis[i]) ans = max(ans, dfs(i));

        if(ans == 0) return -1;
        return ans;
    }
};
103925