- Total Accepted: 163106
- Total Submissions: 417745
- Difficulty: Easy
- Contributors: Admin
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
.
分析
使用双指针, pre 和 cur,非递归
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
|
/**
* 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* pre = head;
ListNode* cur = head->next;
while
(cur != NULL){
if
(cur->val == pre->val){
ListNode * tmp = cur;
cur = cur->next;
pre->next = cur;
delete
tmp;
}
else
{
pre = pre->next;
cur = cur->next;
}
}
return
head;
}
};
|
递归法
删除head之后的list中重复元素,
然后再比较,如果head->val == head->next->val
则返回head->next
1
2
3
4
5
6
7
8
9
10
11
12
13
|
class
Solution {
public
:
ListNode* deleteDuplicates(ListNode* head) {
if
(head == NULL || head->next == NULL)
return
head;
head->next = deleteDuplicates(head->next);
if
(head->next->val == head->val){
ListNode* tmp = head;
head = head->next;
delete
tmp;
}
return
head;
}
};
|