如何对 C++ 中使用邻接表实现的图进行排序？

Question

I try to sort it every time that I add a vertex, so I want to compare the added vertex with the previous vertex, putting the higher one in the left, until I get the head vertex pointer.

The problem that I have is that the function that I have to sort the graph some vertex, works a little bit and after it stop in some point but I can't figure it out where the problem is and the compiler doesn't return any error.

This is my function to sort any time that I add a new vertex.

void Graph::sortGraph(Vertex *pVertex){

/*
* Sort a vertex when it is insert into the graph
* Sorts it from highest to lowest
*/
while(pVertex->previous != NULL){ //Stops when the Vertex is sorted
        if(pVertex->power > pVertex->previous->power){ // To order the graph from highest to lowest
            if(pVertex->previous->previous == NULL){//If it is in position 1 and to
                graphHead = pVertex;
                pVertex->previous->next = pVertex->next;
                pVertex->next->previous = pVertex->previous;
                pVertex->next = pVertex->previous;
                pVertex->previous->previous = pVertex;
                pVertex->previous = pVertex->previous->previous;

            }
            else{ //in any other positions
                pVertex->previous->previous->next == pVertex;
                pVertex->next = pVertex->previous;
                pVertex->previous->previous = pVertex; 
                pVertex->previous->next = pVertex->next;
                pVertex->previous = pVertex->previous->previous;
            }

        } 
        pVertex = pVertex->previous; //go backwards because it have to compare the added vertex and the last one 
}

}

Answer 1

This seems wrong:

            pVertex->previous->previous = pVertex;
            pVertex->previous = pVertex->previous->previous;

you set pVertex->previous->previous to pVertex, then you set pVertex->previous to pVertex->previous->previous that is pVertex itself, so pVertex->previous is pVertex it seems that you create a loop, but maybe I miss something.