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.
题目意思为删除链表中反复的元素
思路: 遍历链表。用两个指针。一个指向前一个。一个指向后一个,当两个所指向的值不相等时,同一时候往后移。当两个指向的值相等时,须要删除一个元素(后一个指针指向的值),然后后一个往后遍历
代码例如以下:
<span style="font-size:18px;">/**
* 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==NULL||head->next==NULL)
return head;
ListNode *p,*q,*pur;
p=head;
q=p->next;
while(q!=NULL)
{
if(p->val==q->val)
{
pur=q;
q=q->next;
p->next=q;
delete pur;
}
else
{
p=q;
q=q->next;
}
}
return head;
}
};</span>