题目描述

在一个排序的链表中，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

解决方案

/* struct ListNode { int val; struct ListNode *next; ListNode(int x) : val(x), next(NULL) { } }; */
class Solution {
public:
    ListNode* deleteDuplication(ListNode* pHead)
    {
        if(!pHead || !pHead->next) return pHead;
        ListNode* start = new ListNode(0);
        start->next = pHead;
        ListNode* pre = start;
        while(pre->next)
        {
            ListNode* cur = pre->next;
            while(cur->next && cur->next->val == cur->val)
            {
                cur = cur->next;
            }
            if(cur != pre->next) 
            {
                pre->next = cur->next;
            }else{
                pre = pre->next;
            }
        }
        return start->next;
    }
};