题解 | #删除链表中重复的结点#

三个指针分别指向前中后三个节点，如果中间节点与后面节点的值相等，就开一个循环把其中相等的值全删掉。

class Solution {
public:
    ListNode* deleteDuplication(ListNode* pHead) {
        if (!pHead || !pHead->next) return pHead;
        ListNode* dummyhead = new ListNode(0), *bef = dummyhead;
        dummyhead->next = pHead;
        ListNode* now = pHead, *nxt = pHead->next;
        while (nxt) {
            if (now->val == nxt->val) {
                while (nxt && now->val == nxt->val) {
                    nxt = nxt->next;
                }
                bef->next = nxt;
                if (!nxt) break;
                now = nxt;
                nxt = nxt->next;
            }
            else {
                bef = now;
                now = nxt;
                nxt = nxt->next;
            }
        }
        return dummyhead->next;
    }
};