Type：medium
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
Example 1:
Input:1->2->3->3->4->4->5Output:1->2->5
Example 2:
Input:1->1->1->2->3Output:2->3
给定一个排好序的链表，删去出现超过一次的所有节点。
判断head及head->next是否为空，若为空，返回head，结束。
首先设一个空的dummy节点指向head，用pre表示想要保留的节点，初始等于dummy，pre->next指向想要保留但未确定是否只出现一次的节点。cur指向当前遍历的节点。
当cur->next的val等于cur->val，cur向后遍历，直至cur-next-val不等于cur-val停止，此时判断pre-next是否与cur是同一个指针，若不等则pre-next指向cur-next，若相等则pre等于pre-next。最后返回dummy-next
ListNode* deleteDuplicates(ListNode* head) {
if(!head || !head->next) return head;
ListNode *dummy = new ListNode(-1);
dummy->next = head;
ListNode *pre = dummy;
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 dummy->next;
}
