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复制链接列表并返回指向新列表的指针

更新时间:2023-12-05 15:58:46

我尚未测试过,但这应该可以完成工作:

I haven't tested it but this should do the job :

Node *copy(Node *list) {
  int count = 0;
  Node *previous = NULL ;
  Node *firstnode = NULL ;

  while (list != NULL){
   Node *newnode = malloc(sizeof(node)) ;

   if (firstnode == NULL)
     firstnode = newnode ;

   memcopy(newnode, list, sizeof(Node)) ;
   newnode->next = NULL ;

   if (previous != NULL)
     previous->next = newnode ;

   previous = newnode ;
   count++;
   list = list->next;
  }

  return firstnode ;
}