在为动态节点列表分配新节点时遇到麻烦,实际上我只能在那里获得第一个节点。一旦检测到重复,要么我没有正确打印它们(不太可能),要么它们没有分配。除插入到队列中的第一个数字外,所有内容打印正常。
typedef struct lineList
{
int lineNum;
LIST *next;
}LIST;
typedef struct nodeTag{
char data[80];
LIST *lines;
struct nodeTag *left;
struct nodeTag *right;
} NODE;
调用addtolist
mover = (*root)->lines;
printf("Node already in the tree!n");
while(mover)
mover = mover->next;
mover = addToList(); //allocate memory for new node
mover->lineNum = line; //set data
添加到列表
LIST *addToList()
{
LIST *pnew;
pnew = (LIST *) malloc(sizeof (LIST)); //memory for LIST
if (!pnew)
{
printf("Fatal memory allocation error in insert!n");
exit(3);
}
pnew->next = NULL; //set next node to NULL
return pnew;
}
树输出到文件(我在这里也有一些节点的小问题打印两次)
void treeToFile(NODE *root, FILE *fp)
{
if(root->left)
{
treeToFile(root->left, fp);
fprintf(fp, "%-15s", root->data);
printList(root->lines, fp);
fprintf(fp, "n");
}
if(root->right)
{
treeToFile(root->right, fp);
fprintf(fp, "%-15s", root->data);
printList(root->lines, fp);
fprintf(fp, "n");
}
return;
}
打印列表
void printList(LIST *myList, FILE *fp)
{
LIST *mover;
mover = myList;
while(mover) //while mover
{
fprintf(fp, "%5d", mover->lineNum); //line nums where string occurs
mover = mover->next; //move to next node
}
}
你失去了链接,看看这段代码:
while(mover)
mover = mover->next;
mover = addToList(); //allocate memory for new node
mover->lineNum = line; //set data
你的时间毫无意义…当'mover' == NULL时,你离开while,你要做的是保留最后一个节点(下一个节点为NULL)
将代码更改为以下
// the IF below is for the case when the queue is empty, so you won't try to dereference a NULL
// in the while condition
if(mover)
while(mover->next)
mover = mover->next;
if(mover) // make sure you have at least one element in the queue
{
mover->next = addToList(); //allocate memory for new node
mover->next->lineNum = line; //set data
}
else // if the queue is empty, then lines will return NULL and you
//are inserting the first element
{
mover = addToList();
mover->lineNum = line;
(*root)->lines = mover; // here you are putting the new element in the first position
// of your queue (It is necessary to do this because
// it is currently empty!
}
上面的代码是修复代码的方法。我有两个建议,这取决于你的队列意味着什么。
第一:这个队列需要排序吗?元素插入的顺序对你有影响吗?
如果是,那么一个队列就是你所需要的,而不是每次插入都要遍历整个列表,你可以有一个这样的结构体:
struct queue
{
LIST *first;
LIST *last;
}
,您将在last->next中插入新元素(第一个元素是不同的,因为last在那里将为空…你需要让next和first指向那个元素)
或者,如果顺序无关紧要,只需在列表的开头添加新元素
mover = addToList();
mover->lineNum = line;
mover->next = *(root)->lines; //supposing you are using the right side correctly in your code,
// you are adding the current list as being the next of your new element
*(root)->lines = mover; //here you are saying that your list starts now at your new element