When analyzing the graph, it is crucial to traverse through all nodes while maintaining a visited array to keep track of visited nodes. If you encounter a node that has already been visited, then there is a cycle present in the directed graph.
For reference and ease of understanding, provided below is a runnable code snippet which can serve as guidance for developing your algorithm.
I have made some enhancements to my previous answer by iterating through all nodes, even covering unreachable paths. Thank you everyone for your contributions :)
array = {};
/*
Assuming
a = 0
b= 1
c = 2
d = 3
e = 4
*/
// Constructing an adjacency list of directed nodes based on the given diagram
array[0] = [4];
array[1] = [4,0];
array[2] = [0,1];
array[4] = [3];
array[3] = [1];
visited = {};
visited[0] = 0;
visited[1] = 0;
visited[2] = 0;
visited[3] = 0;
visited[4] = 0;
list_for_cycle_nodes = [];
for(var node = 0; node<5; node++) {
if (dfs(node)) {
for (var index = 0; index < list_for_cycle_nodes.length; index++) {
console.log(list_for_cycle_nodes[index]+" ");
}
console.log('There is a cycle');
} else {
console.log('Yipee, there is no cycle');
}
}
function dfs(s) {
if(visited[s] == 1) {
return true;
}
list_for_cycle_nodes.push(s);
visited[s] = 1;
var flag = false;
if(array[s].length <= 0) {
return false;
}
for(var index = 0; index<array[s].length; index++) {
flag = flag | dfs(array[s][index]);
if(flag) {
return true;
}
}
return flag;
}
I trust this information proves beneficial in your analysis!