[LeetCode] 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.

 

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思路：和Remove Duplicates from Sorted List一样，双指针，head指向要插入node的前一个node，p指向当前处理的node。

如何判断是否重复呢，判断p和p->next是否一样，若一样，则不插入，不一样就要插入，但是这里忽略了NULL的处理，如果p->next== NULL，要单独处理。。

另外，在找到p->val == p->next->val 时，要找打下一个p，不等于之前的val。

/**
 * 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 NULL;

            ListNode * p = head;
            ListNode dummy(-1);
            head = &dummy;

            while(p)
            {   
                if(p->next == NULL || p->val != p->next->val)
                {   
                    head->next = p;
                    head = p;
                    p = p->next;
                    // to break old link
                    head->next = NULL;
                }   
                else
                {   
                    //find the node which its value is not equla to p->val
                    // and store the node as new p;
                    p = p->next;
                    while(p)
                    {   
                        if(p->next == NULL)
                        {   
                            //should end
                            head->next = NULL;
                            p = NULL;// jump the outer loop
                            break;// jump the inner loop
                        }

                        if(p->val != p->next->val)
                        {
                            p = p->next;
                            break;
                        }
                        else
                            p = p->next;
                    }
                }

                #if 0
                cout << "p->val\t" << p->val << endl;
                cout << "head->val\t" << head->val << endl;
                #endif
            }
            return dummy.next;
        }
};