java list remove duplicates

Given a sorted linked list, delete all nodes that has duplicate numbers, leaving onlydistinct numbers from the O Riginal list.For example,Given 1->2->3->3->4->4->5 , return 1->2->5 .Given 1->1->1->2->3 , return 2->3 .This topic requires the deletion of all occurrences of elements in the list. The code is as follows:listnode* deleteduplicates (listnode* head) {ListNode tmphead (-1); ListNode *ptail = tmpHead

Compared with I actually very different, but very similar to linkedlist reverse, constantly look at pre.next things, I think for a long time, began to use three pointers, completely confused, or back to the original code, just understand Public classSolution { PublicListNode deleteduplicates (ListNode head) {if(head==NULL|| head.next==NULL)returnHead; ListNode h=NewListNode (-1); H.next=Head; ListNode Pre= h, cur = head;//move with cur, the pre runs through the

Given a sorted linkedList, delete AllDuplicates such that all element appear only once.For Example,given1 -1 -2,return 1 -2.Given1 -1 -2 -3 -3,return 1 -2 -3.For this delete list of duplicate elements but for repeating elements need to keep one, the previous encountered the repetition of the element is in front or in the middle of the problem is not, then execute the delete.Runtime:16ms/** * Definition forsingly-linked

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 .Idea: It is a sequential traversal, and then a few pointers are manipulated. A fakehead or Dummynode is added here for easy handling.Boundary Situation!!! This is really a problem, found that a lot of algorithms error is the boundary. Thi

Title Description: (link)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:Recursive version:1 /**2 * Definition for singly-linked list.3 * struct ListNode {4 * int val;5 * ListNode *next;6 * ListNode (int x): Val (x), Ne

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 .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 classSolution {Ten Public: Onelistnode

Question:Given a sorted linked list, delete all nodes that has 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.algorithm:request a memory before the head node, with two nodes pointing to the repeating head and tail .Accepted Code:/** * Definition For singly-linked list. * struct ListNode {* int val; *

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 .1 /**2 * Definition for singly-linked list.3 * public class ListNode {4 * int val;5 * ListNode Next;6 * ListNode (int x) {val = x;}7 * }8 */9 Public classSolution {Ten PublicListNode de

return 1->2return 1->2->3.This problem examines the operation of the list, not difficult/*** Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int X) {val = x;}}*/ Public classSolution { PublicListNode deleteduplicates (ListNode head) {ListNode dummy=NewListNode (0); Dummy.next=Head; ListNode DP=Dummy.next; ListNode previous=NULL; while(dp!=NULL){

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 ./*** Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int X) {val = x;}}*/ Public classSolution {//It's hard to feel yourself, the main breakt

First we need to get the head node, use a Boolean value to mark whether we find a new head or not. When we traverse the linked list,if (cur.val! = Cur.next.val flag) {Newhead = cur;Flag = false;}Once we find the first non-duplicative node, set that node to being head, and flag is false, meaning that we already find the Head node, no need to find a new head anymore.Use a pointer to record the previous node and ahead of the duplicative nodes, use the w

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 .Analysis: Compares the previous number to the last number, if it is different and the number of occurrences is 1, it is added to the result. Pay particular attention to the last node: if it is the same as the previous node, you do not hav

Given a sorted linked list, delete all nodes that have 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. Difficulty: 70. For reference, we need to point the precursor pointer to the previous non-repeating element. If a non-repeating element is found, the precursor pointer knows the element; otherwise, the element is deleted. The algorithm only

/** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode*deleteduplicates (ListNode *head) {ListNode*head = (ListNode *) malloc (sizeof(ListNode)); ListNode*tail = HEAD,*P1 =Head; while(p1) { while(P1->nextp1->val = = p1->next->val) P1 = p1->next;//Go heavyTail->next =P1; Tail= tail->Next; P1= p1->Next; } returnHead->Next; }};

The application of stepper motor thought: You can take one step at a time, or you can walk more than one step at a time (across multiple things with the same attributes). 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 = = NULL)returnHead ListNode *p1, *P2;

Topic: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 .Ideas: Packagelist; Public classremoveduplicatesfromsortedlistii { PublicListNode deleteduplicates (ListNode head) {ListNode P=NewListNode (0); P.next=NULL; ListNode prev=p; ListNode Q=Head; while(Head! =NULL) { while(Q

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 .public class Solution {public ListNode deleteduplicates (ListNode head) { if (head = = NULL | | head.next = = NULL) return head; ListNode temp = new ListNode (0); Temp.next = head; Head = temp;

Problem Description:Given a sorted linked list, delete all nodes that has duplicate numbers, leaving onlydistinct numbers from the O Riginal list.For example,Given 1->2->3->3->4->4->5 , return 1->2->5 .Given 1->1->1->2->3 , return 2->3 .Basic ideas:Sets whether a flag is a variable of repeating number isrepeat, and the current and the next one will isrepeat true. The Isrepeat node is then deleted, leaving the node distinct.Code:ListNode *deleteduplica

/**83. Remove duplicates from Sorted list * @param head * @return linked list to heavy /public ListNode deleteduplicates ( ListNode Head] { if (head = = null) {return null; } ListNode ret = new ListNode ( -1); Ret.next = head; ListNode pre = head;

/** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution {//On my own Public: ListNode* Deleteduplicates (listnode*head) {ListNode* cur=Head; ListNode* ret1=Head; while(cur) { while(Cur->next!=null cur->val==cur->next->val) {cur=cur->Next; } Ret1->next=cur->Next; Ret1=cur->Next; Cur=cur->Next; } returnHead; }};[Leetcode] 83.