16th & 17 remove duplicates from sorted list

Remove duplicates from sorted list 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.

Solution:

<span style="font-weight: normal;">/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode deleteDuplicates(ListNode head) {        if(head==null || head.next==null) return head;        ListNode prev = head, cur = head.next;        while(cur!=null){            if(prev.val==cur.val){//remove cur from the list                prev.next=cur.next;                 cur=cur.next;            }            else{                prev=prev.next;                cur=cur.next;            }        }        return head;            }}</span>

Remove duplicates from sorted List 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.

<span style="font-size:18px;">/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode deleteDuplicates(ListNode head) {        if(head==null || head.next==null) return head;        ListNode newHead = new ListNode(0);//create a new empty head, the value of the new head will not be counted in        newHead.next=head;        ListNode prev = newHead, cur=head;        while(cur!=null&&cur.next!=null){            if(cur.val!=cur.next.val){                prev=prev.next;                cur=cur.next;            }else{                while(cur.next!=null&&cur.next.val==cur.val)    cur=cur.next;                //remove nodes from prev.next to cur                prev.next=cur.next;                cur=cur.next;            }        }        return newHead.next;    }}</span>

Note that the while condition must be cur! = NULL and cur. Next! = NULL. When only one element is left in the traversal list (cur. Next = NULL), it is clear that there is no possibility of equality.

16th & 17 remove duplicates from sorted list