链表的合并哪里不对，求解。该如何处理

2012-04-18

链表的合并哪里不对，求解。C/C++ codevoid merge(node* p1, node* p2){if (p1 NULL || p2 NULL)retur

链表的合并哪里不对，求解。

C/C++ code

void merge(node* p1, node* p2){    if (p1 == NULL || p2 == NULL)        return;    node* head = NULL;    while (p1 && p2)    {            if (p1->data <= p2->data)        {            head = p1;            p1 = p1->next;        }        else        {            head = p2;            p2 = p2->next;        }    }    while (p1 != NULL)    {        head->next = p1;        p1 = p1->next;    }    while (p2 != NULL)    {        head->next = p2;        p2 = p2->next;    }}

[解决办法]
第一个while循环，head指来指去的，到底想指哪里
这里应该是一个if，而不是while,确定头节点
另外，合并后的链表的头节点怎么输出？

C/C++ code

    node* head = NULL;    if (p1 && p2) {        if (p1->data <= p2->data)        {            head = p1;            p1 = p1->next;        }        else        {            head = p2;            p2 = p2->next;        }    }    node* tail = head;    while (p1 && p2)    {            if (p1->data <= p2->data)        {            tail->next = p1;            p1 = p1->next;        }        else        {            tail->next = p2;            p2 = p2->next;        }        tail = tail->next;    }    while (p1 != NULL)    {        tail->next = p1;        tail = tail->next;        p1 = p1->next;    }    while (p2 != NULL)    {        tail->next = p2;        tail = tail->next;        p2 = p2->next;    }
[解决办法]
void merge(node* p1, node* p2)
{
   node* head = NULL,*pt = NULL;

   if (p1 == NULL || p2 == NULL)
       return;

if (p1->data <= p2->data)
   {
       head = p1;
pt = head;
       p1 = p1->next;
   }
   else
   {
       head = p2;
pt = head;
       p2 = p2->next;
   }

   while (p1 && p2)
   {   
       if (p1->data <= p2->data)
       {
           pt->next = p1;
pt = pt->next;
           p1 = p1->next;
       }
       else
       {
           pt->next = p2;
pt = pt->next;
           p2 = p2->next;
       }
   }
   while (p1 != NULL)
   {
       pt->next = p1;
pt = pt->next;
       p1 = p1->next;
   }
   while (p2 != NULL)
   {
       pt->next = p2;
pt = pt->next;
       p2 = p2->next;
   }
}

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