Remove Duplicates from Sorted List II
题目
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.
思路
定义一个前结点p=&dummy,然后分情况讨论cur和cur->next。
1）若cur->next=NULL,则p->next=cur;p=p->next;cur=cur->next
2）cur->next!=NULL,cur->next->val==cur->val,cur移动到不相等的位置。
3）cur->next!=NULL,cur->next->val!=cur->val,操作如1）。
解题
c++版
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* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
if(!head || !head->next)
return head;
ListNode dummy(-1);
ListNode *p=&dummy;
ListNode *cur=head,*q;
while(cur){
if(!cur->next){
p->next=cur;
p=p->next;
cur=cur->next;
}
else{
if(cur->next->val==cur->val){
q=cur->next;
while(q and q->val==cur->val){
q=q->next;
}
cur=q;
}
else{
p->next=cur;
p=p->next;
cur=cur->next;
}
}
}
p->next=NULL;
return dummy.next;
}
};
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Python版
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class Solution:
def deleteDuplicates(self, head):
dummy=ListNode(-1)
if not head or not head.next:
return head
p=dummy
cur=head
while cur:
if not cur.next:
p.next=cur
p=p.next
cur=cur.next
else:
if cur.val!=cur.next.val:
p.next=cur
p=p.next
cur=cur.next
else:
q=cur.next
while q and q.val==cur.val:
q=q.next
cur=q
p.next=None
return dummy.next
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