c - 如何访问堆上的数组链表？

Question

我在访问节点的子数组时遇到问题。我写了两个结构，其中一个包含另一个。我无法访问超出子数组第一个节点的内容。

struct node{
    int distance;
    int destination;
    int weight;
    node *adj;
}; 

struct adjList{
    struct node *node;
    adjList *array;
};// adjList a is made out of an array of "nodes". Essentially each element in the adjList  a should have a pointer to a subarray of "nodes" that i can access.


a=(adjList*) malloc(numOfNodes * sizeof(struct adjList));//allocate space for array of linked lists


for(int j=0; j<numOfNodes; j++){
    array[j].node=malloc(numOfEdges * sizeof(struct node));//allocate space for each linked list in the array
                }


  for(int j=0; j<numOfNodes; j++){
   a[j].node->adj[j]=NULL; //trying to set the "jth's" element of the adjacencylist's "jth" node. This syntax does not work as the compiler wont let me even use it.
      }  

我的整个目标是拥有一个链表数组。不确定为什么此方法不起作用。

Answer 1

要拥有链表数组，您需要创建一个指向链表第一个节点的指针数组。

struct node **array = malloc(sizeof(struct node*) * arraySize /* numOfNodes */);

现在 array[i] 将指向 ith<​​ 链表。

for(int i=0; i<arraySize ; i++){
    struct node *head = NULL;
    /* Allocate nodes for ith linked list */
    for(int j=0; j<numOfNodes; j++) {
        if(0 == j) {
            array[i] = malloc(sizeof(struct node)); //First node of ith linked list
            memset(array[i], 0, sizeof(struct node)); //OR you can use calloc. Required to remove junk pointers in node.
            head = array[i];
        } else {
            head->adj = malloc(sizeof(struct node)); /* Allocate jth node */
            memset(head->adj, 0, sizeof(struct node)); //OR you can use calloc. Required to remove junk pointers in node.
            head = head->adj;
        }
    }
}

你可以遍历 ith<​​ 个链表，如下所示。

struct node *head = array[i];
while(head) {
    printf("\ndist %d dest %d weight %d\n", head->distance, head->destination, head->weight);
    head = head->adj;
}