[802. Find Eventual Safe States](https://leetcode.com/problems/find-eventual-safe-states/) ### 題目描述 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. ### 範例 **Example 1:** ![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/03/17/picture1.png =60%x) ``` 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` <= 10^4^ * 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 * 10^4^]. ### 解答 #### C++ ``` cpp= class Solution { public: vector<int> eventualSafeNodes(vector<vector<int>>& graph) { int numNodes = graph.size(); vector<int> indegree(numNodes); vector<vector<int>> adj(numNodes); for (int nodeID = 0; nodeID < numNodes; nodeID ++) { for (int next : graph[nodeID]) { adj[next].push_back(nodeID); indegree[nodeID] ++; } } queue<int> frontier; for (int nodeID = 0; nodeID < numNodes; nodeID ++) { if (indegree[nodeID] == 0) { frontier.push(nodeID); } } vector<bool> safeNodes(numNodes); while (not frontier.empty()) { int u = frontier.front(); frontier.pop(); safeNodes[u] = true; for (const auto& next : adj[u]) { indegree[next] --; if (indegree[next] == 0) { frontier.push(next); } } } vector<int> ans; for (int nodeID = 0; nodeID < numNodes; nodeID ++) { if (safeNodes[nodeID]) { ans.push_back(nodeID); } } return ans; } }; ``` > [name=Jerry Wu][time=12 Jul, 2023] #### Python ```python= class Solution: def eventualSafeNodes(self, graph: List[List[int]]) -> List[int]: n = len(graph) cache = {} def dfs(node): if node in cache: return cache[node] cache[node] = False for child in graph[node]: if not dfs(child): return False cache[node] = True return True ans = [] for i in range(n): if dfs(i): ans.append(i) return ans ``` > [name=Yen-Chi Chen][time=Thu, Jul 13, 2023] ### Reference [回到題目列表](https://hackmd.io/@Marsgoat/leetcode_every_day)