1、不保留重复元素也就是 2 2 3 4 5 5 5 -> 3 4
可能删掉头节点, 这种问题设置一个虚拟头节点,可以方便不用处理头节点,
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* deleteDuplication(ListNode* head) {
ListNode* L = new ListNode(-1);
L->next=head;
ListNode* p = L;
while(p->next) {
ListNode* q = p->next;
while(q && q->val == p->next->val) q = q->next;
if(p->next->next == q) p=p->next;
else p->next = q;
}
return L->next;
}
};
2、保留重复元素 2 2 3 4 5 5 5 -> 2 3 4 5
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
ListNode* L = new ListNode(-1);
L->next=head;
ListNode* p = head;
while(p) {
ListNode* q = p->next;
while(q && q->val == p->val) q = q->next;
p->next=q;
p=p->next;
}
return head;
}
};