802. Find Eventual Safe States
Đề Bài
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.length1 <= n <= 1040 <= graph[i].length <= n0 <= graph[i][j] <= n - 1graph[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].
Thuật Toán & Kỹ Thuật
⏱️ Thời gian
O(V+E)
💾 Không gian
O(V)
Lời Giải
C++
0802-find-eventual-safe-states.cpp
#define WHITE 0
#define GRAY 1
#define BLACK 2
class Solution {
private:
vector<vector<int>> graph;
vector<int> color;
int n;
public:
bool dfs(int u) {
if (color[u] == BLACK) {
return true;
}
if (color[u] == GRAY) {
return false;
}
color[u] = GRAY;
for (int v : graph[u]) {
bool check = dfs(v);
if (!check) {
return false;
}
}
color[u] = BLACK;
return true;
}
vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
this->graph = graph;
this->n = (int)graph.size();
color.resize(n, WHITE);
for (int i = 0; i < n; ++i) {
if (graph[i].size() == 0) {
color[i] = BLACK;
}
}
for (int i = 0; i < n; ++i) {
if (color[i] == WHITE) {
dfs(i);
}
}
vector<int> ans;
for (int i = 0; i < n; ++i) {
if (color[i] == BLACK) {
ans.push_back(i);
}
}
return ans;
}
};