Given a sorted linked list, delete all duplicates such that each element appear only
once.
For example,
Given 1->1->2
, return 1->2
.
Given 1->1->2->3->3
, return 1->2->3
.
solution:
设定两个节点p和q,其初始值分别热头结点和头结点的下一个节点,当p的节点值等于q的节点值时,q后移一位,同时删除q之前的节点。
当p,q不等时,p、q都后移一位。因为最终循环终止的条件是q->next != NULL,因此在循环体外还要判断p、q手相等。
code:
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
if(head == NULL)
return head;
ListNode* flag = head;
ListNode* p = head->next;
while(p != NULL)
{
if(flag->val != p->val)
{
flag->next = p;
flag = p;
}else{
flag->next = NULL;
}
p = p->next;
}
return head;
}
};