很简单的一题,需要注意的是如果某结点重复了记得将其删除。

ListNode *deleteDuplicates(ListNode *head)
{ if (head == nullptr)
return nullptr;
ListNode *prev = head;
for (ListNode *cur = head->next; cur != nullptr; cur = cur->next)
{
if (cur->val == prev->val)
{
prev->next = cur->next; delete cur;
}
else
prev->next = cur;
}
}

这题采用递归来做,当两结点不等时,注意递归调用的形式。

ListNode *deleteDuplicates2(ListNode *head)
{
if (!head || !head->next)return head;
ListNode *p = head->next;
if (p == head)
{
while (p&&p==head)
{
ListNode *temp = p;
p = p->next;
delete temp; }
delete head;
return deleteDuplicates2(p);
}
else
{
head->next=deleteDuplicates2(head->next);
return head;
}
}

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