1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
|
// 802. Find Eventual Safe States
#include "leetcode.h"
/*
* 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.
*/
class Solution {
public:
vector<int> eventualSafeNodes(vector<vector<int>> &graph) {
int n = graph.size();
vector<vector<int>> g(n);
vector<int> dp(n, 0);
for (int i = 0; i < n; i++)
for (int j = 0; j < graph[i].size(); j++) {
g[graph[i][j]].push_back(i);
dp[i]++;
}
queue<int> q;
for (int i = 0; i < n; i++)
if (!dp[i])
q.push(i);
vector<int> ans;
while (!q.empty()) {
int u = q.front();
q.pop();
ans.push_back(u);
for (int v : g[u])
if (!--dp[v])
q.push(v);
}
sort(ans.begin(), ans.end());
return ans;
}
};
int main() {
Solution obj;
vvd(int) g1 = {{1, 2}, {2, 3}, {5}, {0}, {5}, {}, {}};
vvd(int) g2 = {{1, 2, 3, 4}, {1, 2}, {3, 4}, {0, 4}, {}};
for (auto i : obj.eventualSafeNodes(g1))
printf("%d ", i); // expect: 2 4 5 6
printf("\n");
for (auto i : obj.eventualSafeNodes(g2))
printf("%d ", i); // expect: 4
printf("\n");
}
|