删除链表中重复的结点

Posted 2022-11-23 ssopp24

ListNode* deleteDuplication(ListNode* pHead)
    
        if ( NULL == pHead || NULL == pHead->next )
            return pHead;
        
        ListNode* p1 = pHead;
        ListNode* p2 = pHead->next;
        ListNode* pPrev = pHead;
        ListNode* pNewHead = pHead;
        
        if ( p1->val == p2->val )
            pPrev = NULL;
            pNewHead = NULL;
        
        
        while ( 1 )
            if ( p1->val == p2->val )//在这个大逻辑中，删除节点
                while ( p1->val == p2->val )
                    if ( NULL == p2->next )
                        if ( NULL == pNewHead )
                            return NULL;
                        else
                            pPrev->next = NULL;

                            delete p1;
                            delete p2;

                            return pNewHead;
                        
                    
                    else
                        ListNode* tmp = p2;
                        p2 = p2->next;

                        delete tmp;
                        tmp = NULL;
                    
                
                
				if ( NULL == pNewHead )
					pNewHead = pPrev = p2;
				
				delete p1;
				p1 = p2;

				if ( NULL == p2->next )
					if ( pPrev != p2 )
						pPrev->next = p2;

					return pNewHead;
				
				else
					p2 = p2->next;

					if ( p1->val == p2->val )
						if ( p1 == pNewHead )
							pPrev = NULL;
							pNewHead = NULL;
						
					
				
			
			else
				if ( p1 != pPrev )
					pPrev->next = p1;

				pPrev = p1;
				p1 = p1->next;

				if ( NULL != p2->next )
					p2 = p2->next;
				else
					pPrev->next = p2;
					return pNewHead;
				
			
        
    

在一个有序的链表中，存在重复的结点，请删除该链表中重复的结点，重复的结点不保留，返回链表头指针。 

例如，链表1->2->3->3->4->4->5 处理后为 1->2->5