java list remove duplicates

"083-remove duplicates from Sorted list (to remove duplicate nodes in a sorted single-linked list)""leetcode-Interview algorithm classic-java Implementation" "All topics Directory Index"Original QuestionGiven a sorted linked

) cur =cur.next;///First, look for the node with the first node value and the node value different from the current table head node Head.next=deleteduplicates (cur);//when found, take the second step, that is, the current header node next point to the node you just found. Here is a recursive call, the above found on the non-repeating node is cur, then this recursive return is exactly the cur, and at the same time to perform the cur as a nod to workreturnHead ;//Return to First node}}Iterative so

ListNode (3);p 2.next = p3; ListNode P4 = new ListNode;p 3.next = P4;prinf (head);p Rinf (Dp.deletedup (Head));} private static void Prinf (ListNode input) {while (input!=null) {System.out.print (input.val+ "); input = Input.next;} System.out.println ();} Public ListNode Deletedup (ListNode head) {if (head==null| | Head.next==null) return head;//if No head, what is should I do? ListNode p=head;int i=0;//system.out.println (p.val+ "and" +p.next.val); while (p.next! = null) {if (p.val==p.next.val

Title: http://oj.leetcode.com/problems/remove-duplicates-from-sorted-list/ Given a sorted linked list, delete all duplicates such which each element is appear only once. For example,Given 1->1->2, return 1->2.Given 1->1->2->3->3, return 1->2->3. Title translation: Given an

Given a sorted linked list, delete all duplicates such this each element appear only once.For example,Given 1->1->2 , return 1->2 .Given 1->1->2->3->3 , return 1->2->3 .Idea: Define two pointers, pre and cur, if cur and pre are equal, then cur move, if not equal, move at the same time.public class Solution {public ListNode deleteduplicates (ListNode head) { if (head==null| | Head.next==null)//Fir

2.next = p3; ListNode P4 = new ListNode;p 3.next = P4;prinf (head);p Rinf (Dp.deletedup (Head));} private static void Prinf (ListNode input) {while (input!=null) {System.out.print (input.val+ "); input = Input.next;} System.out.println ();} Public ListNode Deletedup (ListNode head) {if (head==null| | Head.next==null) return head;//if No head, what is should I do? ListNode p=head;int i=0;//system.out.println (p.val+ "and" +p.next.val); while (p.next! = null) {if (p.val==p.next.val p.next.next!=n

unification of the two. Also highlighted in the relevant documentation.Solution One:1 Public Static BooleanContainsNearbyAlmostDuplicate1 (int[] Nums,intKintt) {2 if(kreturn false;3 4TreesetNewTreeset();5 for(inti=0;i){6 intCurrent =Nums[i];7 if(Set.floor (current)! =NULL current8(Set.ceiling (current)! =NULL) current>=set.ceiling (current)-t)9 return true;Ten Set.add (current); One A if(i>=k) -Set.remov

Pre.next.val = =head.val) { Do{Head=Head.next; } while(head!=NULL Head.val = =pre.next.val); Pre.next=Head; } ElsePre =Pre.next; } returnreq; }}Modify the idea: You can create a virtual table header, next point to head, so that the original code in the initial part of the extra judgment to optimize the code is very concise.Modified code: Public classSolution { PublicListNode deleteduplicates (ListNode head) {ListNode dunmy=NewListNode (I

Test Instructions:Given a sorted linked list, delete all nodes that has duplicate numbers, leaving only distinct numbers from the Original list.For example,Given 1->2->3->3->4->4->5 , return 1->2->5 .Given 1->1->1->2->3 , return 2->3 .Title:Given an ordered list, all duplicate nodes are deleted, and the remainder are the different node elements in the original linked list.Like whatGiven 1->2->3->3->4->4->5,

Test Instructions:Given a sorted linked list, delete all nodes that has duplicate numbers, leaving only distinct numbers from the Original list.For example,Given 1->2->3->3->4->4->5 , return 1->2->5 .Given 1->1->1->2->3 , return 2->3 .Title:Given an ordered list, delete all the repeated nodes. The remainder is not the same node element in the original list.Example.Given 1->2->3->3->4->4->5. Return 1->2->5 .

Given a sorted linked list, delete all nodes that has duplicate numbers, leaving only distinct numbers from the Original list.For example,Given 1->2->3->3->4->4->5 , return 1->2->5 .Given 1->1->1->2->3 , return 2->3 .Problem Solving Ideas:In contrast to I, this problem requires that no duplicate nodes be retained. Therefore, an index record is required at the end of the node that is currently reserved, and then traversed with another node, and the dup

2.1 Write code to remove duplicates from an unsorted linked list.Follow upHow would-solve this problem if a temporary buffer isn't allowed?This problem allows us to remove duplicates from the unordered list, and there are two similar problems in Leetcode that are the

; node = nextNode; } else { node = node.next; } } return head; } } public class Solution { if (head == null) return head; ListNode node = head; while (node.next != null) { if (node.val == node.next.val) { node.next = node.next.next; } els

Label: style blog Io color AR for SP Div on Given a sorted Linked List, delete all duplicates such that each element appear onlyOnce. For example,Given1-> 1-> 2, Return1-> 2.Given1-> 1-> 2-> 3-> 3, Return1-> 2-> 3. Remove the nodes with duplicate values in the incremental linked list as stated in the question.

LeetCode 83 Remove Duplicates from Sorted List (Remove duplicate elements from Sorted List )(*)Translation Given a sorted Linked List, delete all repeated elements so that each element appears only once. For example, if 1-> 1-> 2

} -}Extension questions:Delete a repeating element in a sorted listEasy To remove duplicate elements from the sorted list view Run results Given a sorted list, delete all duplicate elements each element leaves only oneSample ExampleGive 1->1->2->null, return 1->2->nullGive 1->1->2->3->3->null, return 1->2->3->null1 /**2 * Definition for ListNode3 * public c

one. The Code is as follows: /** * 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 || !head->next) return head; ListNode *start = new ListNode(0); start->next = head; ListNode *pre = start; while (pre->next) { ListNode *

Leetcode 2. Remove duplicate elements from the ordered linked list and array Remove Duplicates, leetcodeduplicates1. linked List, so that duplicate elements only appear once Remove Duplicates