remove-duplicates-from-sorted-list I&II——去除链表中重复项

Posted 2020-09-24 鸭子船长

I、Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given1->1->2, return1->2.
Given1->1->2->3->3, return1->2->3.

 

PS：遍历，而后记录pre，并删除后续重复node

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *deleteDuplicates(ListNode *head) {
12         ListNode *cur=head,*pre=NULL;
13         while(cur!=NULL){
14             if(pre==NULL){
15                 pre=cur;
16                 cur=cur->next;
17                 continue;
18             }
19             ListNode *next=cur->next;
20             if(pre->val==cur->val){
21                 pre->next=next;
22             }else
23                 pre=cur;
24             cur=next;
25         }
26         return head;
27     }
28 };

II、

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given1->2->3->3->4->4->5, return1->2->5.
Given1->1->1->2->3, return2->3.

 

删除所有重复项。

PS：循环比较next->val==cur->val，若next跳动则剔除cur至next之间的节点。否则右移left指针。

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *deleteDuplicates(ListNode *head) {
12         if(head==NULL) return NULL;
13         ListNode h(-1);
14         ListNode *res=&h;
15         res->next=head;
16         ListNode *cur=head,*pre=NULL,*left=res;
17         while(cur!=NULL){
18             ListNode* next=cur->next;
19             while(next!=NULL&&cur->val==next->val){
20                 next=next->next;
21             }
22             if(next==cur->next){
23                 left=cur;
24             }else{
25                 left->next=next;
26             }
27             cur=next;
28         }
29         return res->next;
30     }
31 };