[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
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