链表分割--更高难度的场景(自己设想的场景,呵呵)

//现有一链表的头指针 ListNode* pHead，给一定值x，
//编写一段代码将所有小于x的集中于前半部分，大于x的集中于后半部分，
//若链表内有x则居于中间,没有则不显示
//且不能改变原来的数据相对顺序，返回重新排列后的链表的头指针。
//示例 x=5
//1 6 2 7 3 5 9 8  (5)-->1 2 3 5 6 7 9 8
//1 6 2 7 8 5 9 3  (5)-->1 2 3 5 6 7 8 9 
//1 2 3 (5)-->1 2 3
//6 7 8 (5)--->6 7 8
//5 5 5 (5)--->5 5 5 

ListNode* Partition(ListNode* phead,int x)
{
    ListNode* smallhead = NULL;
    ListNode* smallx = NULL;
    ListNode* bighead = NULL;
    ListNode* bigx = NULL;
    ListNode* flaghead = NULL;
    ListNode* flagx = NULL;
    if (phead == NULL)
        return NULL;
    while (phead)
    {
        if (phead->val < x)
        {
            if (smallhead == NULL)
            {
                smallx = phead;
                smallhead = smallx;
            }
            else
            {
                smallx->next = phead;
                smallx = smallx->next;
            }
        }
        else if (phead->val > x)
        {
            if (bighead == NULL)
            {
                bigx = phead;
                bighead = bigx;
            }
            else
            {
                bigx->next = phead;
                bigx = bigx->next;
            }
        }
        else
        {
            if (flaghead == NULL)
            {
                flagx = phead;
                flaghead = flagx;
            }
            else
            {
                flagx->next = phead;
                flagx = flagx->next;
            }
        }
        phead = phead->next;
    }
    if(bigx)
    bigx->next = NULL;
    if (flaghead != NULL)
    {
        if (smallhead == NULL)
        {
            flagx->next = bighead;
            return flaghead;
        }
        else
        {
            smallx->next = flaghead;
            flaghead->next = bighead;
            return smallhead;
        }
    }
    else
    {
        if (smallhead == NULL)
        {
            return bighead;
        }
        else
        {
            smallx->next = bighead;
            return smallhead;
        }
    }

}