Question

I want to make a graph using linked list in the following way. Here the last node's next value is NULL.

我从https://www.geeksforgeeks.org/graph-and-its-representations/获得的程序是：

#include <stdio.h>
#include <stdlib.h>

// A structure to represent an adjacency list node
struct AdjListNode
{
    int dest;
    struct AdjListNode* next;
};

// A structure to represent an adjacency list
struct AdjList
{
    struct AdjListNode *head; 
};

// A structure to represent a graph. A graph
// is an array of adjacency lists.
// Size of array will be V (number of vertices 
// in graph)
struct Graph
{
    int V;
    struct AdjList* array;
};

// A utility function to create a new adjacency list node
struct AdjListNode* newAdjListNode(int dest)
{
    struct AdjListNode* newNode =
     (struct AdjListNode*) malloc(sizeof(struct AdjListNode));
    newNode->dest = dest;
    newNode->next = NULL;
    return newNode;
}

// A utility function that creates a graph of V vertices
struct Graph* createGraph(int V)
{
    struct Graph* graph = 
        (struct Graph*) malloc(sizeof(struct Graph));
    graph->V = V;

    // Create an array of adjacency lists.  Size of 
    // array will be V
    graph->array = 
      (struct AdjList*) malloc(V * sizeof(struct AdjList));

    // Initialize each adjacency list as empty by 
    // making head as NULL
    int i;
    for (i = 0; i < V; ++i)
        graph->array[i].head = NULL;

    return graph;
}

// Adds an edge to an undirected graph
void addEdge(struct Graph* graph, int src, int dest)
{
    // Add an edge from src to dest.  A new node is 
    // added to the adjacency list of src.  The node
    // is added at the begining
    struct AdjListNode* newNode = newAdjListNode(dest);
    newNode->next = graph->array[src].head;
    graph->array[src].head = newNode;

    // Since graph is undirected, add an edge from
    // dest to src also
    newNode = newAdjListNode(src);
    newNode->next = graph->array[dest].head;
    graph->array[dest].head = newNode;
}

// A utility function to print the adjacency list 
// representation of graph
void printGraph(struct Graph* graph)
{
    int v;
    for (v = 0; v < graph->V; ++v)
    {
        struct AdjListNode* pCrawl = graph->array[v].head;
        printf("\n Adjacency list of vertex %d\n head ", v);
        while (pCrawl)
        {
            printf("-> %d", pCrawl->dest);
            pCrawl = pCrawl->next;
        }
        printf("\n");
    }
}

// Driver program to test above functions
int main()
{
    // create the graph given in above fugure
    int V = 5;
    struct Graph* graph = createGraph(V);
    addEdge(graph, 0, 1);
    addEdge(graph, 0, 4);
    addEdge(graph, 1, 2);
    addEdge(graph, 1, 3);
    addEdge(graph, 1, 4);
    addEdge(graph, 2, 3);
    addEdge(graph, 3, 4);

    // print the adjacency list representation of the above graph
    printGraph(graph);

    return 0;
}

我的疑问是：addge（）函数在列表中添加一个节点，并且它的下一个指针永远不会为NULL。但是在printGraph（）中会遍历temp，直到它变为NULL。请澄清我的疑问。我误解了addge（）函数吗？

Answer 1

您共享的代码具有很好的注释并且可以自我解释。但是似乎您不清楚代码的某些部分。让我们检查代码部分。当main()调用createGraph(V);时，它实际上为每个顶点创建了一个空的邻接表。这里的空白列表表示head指针已设置为NULL。

然后addEdge(graph, src, dest);在列表的开头添加一个新节点，这里首先为newNode的链接分配了head指针，然后newNode成为头。因为head最初设置为NULL，所以list的末端节点将始终指向NULL。

在由链表创建的图形中添加边

1 个答案: