LeetCode第19天--链表分割

解题思路：首先，先定义两个新的头节点，分别用来保存小于x的节点和不小于x的节点，遍历原链表，当前节点小于x的时候，将它放在定义的新的节点后，当前节点不小于x的时候，将它放在定义的另一个新的节点后。最后将全部小节点的尾链接上大节点的头即可。画图分析:

代码实现：

ListNode* partition(ListNode* pHead, int x) 
   {
        ListNode* cur = pHead;
        ListNode *smallhead =NULL,*bighead = NULL;
        ListNode *smalltail = NULL,*bigtail = NULL;
        while(cur)
        {
           if(cur->val < x)
           {
               if(smallhead == NULL)
               {
                   smallhead = smalltail = cur;
               }
               else
               {
                   smalltail->next = cur;
                   smalltail = cur;
               }
           }
            else
            {
                if(bighead == NULL)
               {
                   bighead = bigtail= cur;
               }
               else
               {
                   bigtail->next = cur;
                   bigtail = cur;
               }
            }
            cur = cur->next;
        }
        if(smallhead && bighead)
        {
            smalltail->next = bighead;
            bigtail->next = NULL;
        }
        else if(smallhead == NULL && bighead)
        {
            return bighead;
        }
        return smallhead;
    }