Maybee you could try a greedy algorithm:
1. Create a list sortedList that stores each pair of nodes i and j and is sorted by the
weight w(i,j).
2. Create a HashSet connectedNodes that is empty at the beginning
3. while (connectedNodes.size() < n)
element := first element of sortedList
if (connectedNodes.isEmpty())
connectedNodes.put(element.nodeI);
connectedNodes.put(element.nodeJ);
delete element from sortedList
else
for(element in sortedList) //start again with the first
if(connectedNodes.get(element.nodeI) || connectedNodes.get(element.nodeJ))
if(!(connectedNodes.get(element.nodeI) && connectedNodes.get(element.nodeJ)))
//so it does not include already both nodes
connectedNodes.put(element.nodeI);
connectedNodes.put(element.nodeJ);
delete element from sortedList
break;
else
continue;
So I explain step 3 a little bit:
You add as long nodes till all nodes are connected to one other. It is sure that the graph is connected, because you just add a node, if he has a connection to an other one already in the connectedNodes
list.
So this algorithm is greedy what means, it does not make sure, that the solution is optimal. But it is a quite good approximation, because it always takes the shortest edge (because sortedList
is sorted by the weight of the edge).
Yo don't get duplicates
in connectedNodes
, because it is a HashSet
, which also make the runtime faster.
All in all the runtime should be O(n^2) for the sorting at the beginning and below its around O(n^3), because in worst case you run in every step through the whole list that has size of n^2 and you do it n times, because you add one element in each step.
But more likely is, that you find an element much faster than O(n^2), i think in most cases it is O(n).