LeetCode: Remove Duplicates from Sorted List II

思路：这个题目又是缝缝补补过的，利用指针q指示当前待审查元素，p为指向剩下链表（删除重复元素后）的最后一个元素，t为新元素，找到t->val != q ->val 然后将p 指向 t，删除了中间重复的元素，如果t 就是q 指向的元素，那么不用删除，p,q的位置都要移动。

class Solution {public:    ListNode *deleteDuplicates(ListNode *head) {if(head){ListNode *p, *q, *t;p = q = head;t = head->next;bool flag = 0;while(t){if(t->val != q->val){if(head->next != t  && q == head){p = q = head = t;}else{if(t == q->next) p = q;p->next = t;q = t;}flag = 0;}elseflag = 1;t = t->next;}if(q->next)p->next = t;if(q==head && flag == 1) head = NULL;}return head;    }};