题意:删掉单链表里重复的节点,如:
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

思路:真的真的是很简单的题啊,但是有个陷阱,也怪自己练的太少,都忘记了这个。思路其实就是判断如果p->val == p->next->val就删掉p->next节点,如果新的p->next的值与p的值不相等则往后移动。这里的判断条件有两个,

一个是p->val != p->next->val,一个是p->next != NULL。

代码:

ListNode* deleteDuplicates(ListNode* head)
{
ListNode* p = head;
if(head == NULL || head->next == NULL)
return head;
while(p->next != NULL)
{
if(p->val == p->next->val)
{
ListNode* temp = p->next;
p->next = p->next->next;
free(temp);
}
if(p->next != NULL && p->val != p->next->val)
{
p = p->next;
}
}
return head;
}

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