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.

很常规的解法，为去除头节点的特殊性，需要用到虚拟头结点技术。

C++ code
    ListNode *deleteDuplicates(ListNode *head) {
        if (!head) return NULL;

        ListNode *dummyHead = new ListNode(0);
        dummyHead->next = head;

        ListNode *prev = dummyHead, *curr = head->next;
        int curVal = head->val;
        while (curr) {
            if (curr->val == curVal) {
                for (curr = curr->next; curr != NULL && curr->val == curVal; curr = curr->next) ;
                prev->next = curr;
                if (curr) {
                    curVal = curr->val;
                    curr = curr->next;
                }
            }
            else {
                prev = prev->next;
                curVal = curr->val;
                curr = curr->next;
            }
        }

        return dummyHead->next;
    }

LeetCode[Linked List]: Remove Duplicates from Sorted List II