NP-hardness
The NP-hardness of the unweighted longest problem can be shown using a reduction from the Hamiltonian path problem: a graph G has a Hamiltonian path if and only if its longest path has length n − 1, where n is the number of vertices in G. Because the Hamiltonian path problem is NP-complete, this reduction shows that the decision version of the longest path problem is also NP-complete. In this decision problem, the input is a graph G and a number k; the desired output is "yes" if G contains a path of k or more edges, and no otherwise.
If the longest path problem could be solved in polynomial time, it could be used to solve this decision problem, by finding a longest path and then comparing its length to the number k. Therefore, the longest path problem is NP-hard. It is not NP-complete, because it is not a decision problem.
In weighted complete graphs with non-negative edge weights, the weighted longest path problem is the same as the Travelling salesman path problem, because the longest path always includes all vertices.
Read more about this topic: Longest Path Problem