82 Remove Duplicates from Sorted List II

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,

Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

Solution

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if ( head == NULL ) return head;
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        ListNode *ptail = dummy, *p = head;
        int val;
        while (true) {
            if ( p->next == NULL ) { ptail->next = p; break; }
            // note the difference between the following two situations: 
            // duplicate
            if ( p->val == p->next->val ) {
                val = p->val;
                while ( p->val == val ) {
                    p = p->next;
                    if ( p == NULL ) break;
                }
                if ( p == NULL ) { ptail->next = p; break; }
            }
            // vs. non-duplicate
            else {
                ptail->next = p;
                if ( p == NULL ) break;
                ptail = p;
                p = p->next;
            }
        }
        return dummy->next;
    }
};

Notes
Pay attention to a lot of DETAILS!
ptail->next is different for "duplicate" case and "non-duplicate case":

• duplicate case: ptail->next does not necessarily point to the first node after the consecutive duplicate nodes. ptail remains the same, we will examine pointer p in the next loop!

  corner case: [1,2,3,3,4,4,5]
  

  Inappropriate linkage would give the erroneous answer: [1,2,4,4,5].
• non-duplicate case: ptail->next points to the next "unique nodes";

Another solution

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if ( head == NULL or head->next == NULL ) return head;
        ListNode *dummy = new ListNode(0);
        ListNode *tail = dummy;
        ListNode *last = head, *p = head->next;
        while ( p != NULL ) {
            if ( p->val != last->val ) {
                tail->next = last;
                tail = tail->next;
            }
            else {
                while ( p != NULL and p->val == last->val ) p = p->next;
                if ( p == NULL ) { last = NULL; break;}
            }
            last = p;
            p = p->next;
        }
        tail->next = last;
        return dummy->next;
    }
};

Notes
这个解法中注意tail->next = last的处理。