题解 | #合并两个排序的链表#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param pHead1 ListNode类 
 * @param pHead2 ListNode类 
 * @return ListNode类
 */
 typedef struct ListNode listnode;
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2) {
     if(pHead1==NULL || pHead2 ==NULL )
      {
         return (pHead1==NULL)? pHead2 :pHead1;//三目操作符
      }
    listnode*n1,*n2;
    n1 = pHead1;
    n2 = pHead2;
    listnode *phead,*ptail;
    phead = ptail = (listnode*)malloc(sizeof(listnode));//定义哨兵结点
    while(n1 && n2)
    {
        if(n1->val < n2->val)
        {
            ptail->next = n1;
            ptail = ptail->next;
            n1 = n1->next;
        }
        else 
        {
            ptail->next = n2;
            ptail = ptail->next;
            n2 = n2->next;
        }
    }
    if(n1)
    {
        ptail->next = n1;
    }
    if(n2)
    {
        ptail->next = n2;
    }
    listnode*pf = phead->next;
    free(phead);//销毁堆上申请的空间
    phead = ptail = NULL;
    return pf;
}