802. Find Eventual Safe States --- M. !

There is a directed graph of n nodes with each node labeled from 0 to n - 1. The graph is represented by a 0-indexed 2D integer array graph where graph[i] is an integer array of nodes adjacent to node i, meaning there is an edge from node i to each node in graph[i].

A node is a terminal node if there are no outgoing edges. A node is a safe node if every possible path starting from that node leads to a terminal node (or another safe node).

Return an array containing all the safe nodes of the graph. The answer should be sorted in ascending order.

 

Example 1:

Input: graph = [[1,2],[2,3],[5],[0],[5],[],[]]
Output: [2,4,5,6]
Explanation: The given graph is shown above.
Nodes 5 and 6 are terminal nodes as there are no outgoing edges from either of them.
Every path starting at nodes 2, 4, 5, and 6 all lead to either node 5 or 6.

Example 2:

Input: graph = [[1,2,3,4],[1,2],[3,4],[0,4],[]]
Output: [4]
Explanation:
Only node 4 is a terminal node, and every path starting at node 4 leads to node 4.

 

Constraints:

• n == graph.length
• 1 <= n <= 104
• 0 <= graph[i].length <= n
• 0 <= graph[i][j] <= n - 1
• graph[i] is sorted in a strictly increasing order.
• The graph may contain self-loops.
• The number of edges in the graph will be in the range [1, 4 * 104].

A:

就是简单的DFS。

class Solution {

public:

vector<int> eventualSafeNodes(vector<vector<int>>& graph) {

// 0: unchecked, 1: checking, 2: terminal, 3: not terminal

vector<int> isTerminal(graph.size(), 0);

vector<int> res;

for (int i = 0; i < graph.size(); i++) {

if (dfs(graph, isTerminal, i))

res.push_back(i);

}

return res;

}

private:

bool dfs(vector<vector<int>>& graph, vector<int>& isTerminal, int u) {

if (isTerminal[u] == 2) {

return true;

}

if (isTerminal[u] == 3 || isTerminal[u] == 1) {

return false;

}

// now, isTerminal[u] == 0

isTerminal[u] = 1; // checking now

for (auto v : graph[u]) {

if (!dfs(graph, isTerminal, v)) {

isTerminal[u] = 3;

return false;

}

}

isTerminal[u] = 2;

return true;

}

};