c++链表怎么输出

c++链表怎么输出

输出如下：
LinkList::LinkList()
在这个函数中，，建链表的这两句话p->next=Link->next;Link->next=p，用的头插法建链表，意思就是说每次把新的节点p放在链表首部，，所以建好的顺序肯定与输入元素的顺序是相反的.要想按原顺序，，则把上述的这两句改为 ：
p->next=r->next;
r->next=p;
r=p;
即可。
在该函数体内，加一个变量声明 LNode *r=Link; 就好了。改制后运行结果如下：How many node do you want:2
1: 1
2: 2
How many node do you want:2
1: 3
2: 4
How many node do you want:2
1: 5
2: 6
The LinkList is 1 2
The LinkList is 3 4 5 6
Press any key to continue

int print(qq* QQ )
{
qq* p=QQ;
int i=0;
while(p!=NULL)
{
cout<p=p.next;
i++;
}
return i;
}

#include
#include
#define MAX_LEN 10
typedef struct _LINKNODE
{
int value;
struct _LINKNODE *next;
}LINKNODE;

LINKNODE* revertLink(LINKNODE* head)
{
LINKNODE* revert = NULL;
LINKNODE* temp = NULL;
while(head != NULL)
{
temp = head->next;
head->next = revert;
revert = head;
head = temp;
}
return revert;
}
void main()
{
int array[MAX_LEN] = ;
LINKNODE* head = NULL;
LINKNODE* curNode = NULL;
LINKNODE* temp = NULL;
LINKNODE* revert = NULL;
for(int i=0; i {
temp = (LINKNODE*) malloc(sizeof(LINKNODE));
temp->value = array[i];
temp->next = NULL;
if(head == NULL)
{
head = temp;
curNode = temp;
}
curNode->next = temp;
curNode = curNode->next;
}
printf("初始化后的链表是:\n");
curNode = head;
while(curNode != NULL)
{
printf("%d ", curNode->value);
curNode = curNode->next;
}
revert = revertLink(head);

printf("\n逆置后链表是:\n");
while(revert != NULL)
{
printf("%d ", revert->value);
revert = revert->next;
}
printf("\n");
}
看看吧...

我暂时保留我的看法！