picard remove duplicates

Because the list is used, there are several ways to remove duplicate data. Recorded in this ...Test data:listMethod One:hashsetSeveral ways to remove duplicates in the list collection-reproduced

/** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode* Deleteduplicates (listnode*head) { if(Head ==null)returnHead; if(Head->next ==null)returnHead; ListNode*res =Head; intHead_val; while(head) {Head_val= head->Val; if(Head->next head->next->val = =head_val) { while(Head Head->val = = head_val)//Remove the beginning of

do not know the size of the problem, it is very bad. It also caused me to waste a quick 2 hours =.= (listnode* res,*tmp;STD:: mapint,int>Ans ListNode *deleteduplicates (ListNode *head) {ListNode * I,*K,*AA;intJBOOLflag=true;if(head = = NULL | | head->next = = NULL)returnHead for(j=-65536;j65537; j + +) Ans.insert (STD::p airint,int> (J,0)); for(I=head;i!=null;i=i->next) {ans[i->val]++;//cout} for(I=head;i!=null;i=i->next) {if(ans[i->val]==1){int*a =New int; ListNode L (i->val);//cout

found that Prev is Fakenode node initially. Because the Fakenode node initially points to the head node.Runtime:8ms/** * Definition forsingly-linked list. * struct ListNode {*intVal * ListNode *Next; * ListNode (intx): Val (x),Next(NULL) {} * }; */classSolution { Public: listnode* deleteduplicates (listnode* head) {ListNode * fakenode=NewListNode (0); Fakenode->Next=head; ListNode *prev=fakenode; while(head!=NULL) {if(head->Nexthead->val==head->Next->val) {intvalue=head->val; while(Headhead-

; * ListNode *next; * ListNode (int x): Val (x), Next (NULL) {} *}; */class Solution {public:listnode* deleteduplicates (listnode* head) {if (head==null) return NULL; if (Head->next==null) return head; listnode* P=head; listnode* Q=head; listnode* k=p->next; BOOL Flag=false; while (k) {if (p->val==k->val) {if (p==q) {p=q=k; Head=k; k=k->next; } else {listnode* tobedeleted=k; p->next=tobedeleted->next; k=p->next; Delete tobedeleted; Tobedeleted=nu

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

It's easy to repeat the data in the list result, as long as. dinstinct (). However, if you want to remove duplicate data based on a field, the above method will not be helpful. We need to rewrite a method. Let's look at the example directly. [Serializable] public class HomePageUserModel { public int UserID { get; set; } public string TitleUserName { get; set; } public string ShowUserName { get; set; } public DateTim

Tags: leetcode linked list algorithm virtual header Node 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. To remove the special characteristics of the header node, the virtual header node technology is required. C++ code ListNode *deleteDuplicates(ListNode *head) { if (!head)

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.Idea: This question in just start to do when thinking of a bit, how can not be solved correctly. Rewrite all of the code to remove it later. The idea is to record the current node P=head, then head down, and when the head value is not equal to th

LeetCode [Linked List]: Remove Duplicates from Sorted List II, leetcodeduplicates 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. To remove the special characteristics of the header node, the virtual

Title: Give a sorted array, remove duplicate elements in the array and allow up to two elements of the same, and finally return the processed array length, and the array is collated.Algorithm idea: When the array length is less than 3 o'clock do not tidy up the array, directly return the length of the array; When the array length is greater than or equal to 3 o'clock, with the pre record of the precursor element, the flag tag repeats once, p records t

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:Create a new auxiliary header, using two pointers pre,p;Start cycle judgment1. Pre->next=p->next, stating that there is the same element, then p=p->next; know that P->next is empty or discovers new elements;2. Judge P->next!=p, if there is the same e

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 .Delete the duplicate numbers in the list, and note that this is the number that will be removed once the numbers have been duplicated. So how to completely remove is a technical live. Of course, in order to deal with the deletion of the h

Description: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 .Code:1listnode* Deleteduplicates (listnode*head) {2 if(head)3 {4listnode* p =head;5listnode* Lastnode =head;6 7 intLastdeleteval =0;8 BOOLFlag =false;//flag is false to indicate that no elements have been deleted9

The array type does not provide a way to repeat, and if you want to kill the repeating elements of the array, you have to do it yourself:1 function Unique (arr) {2 var result = [], isrepeated; 3 for (var i = 0, len = arr.length; i isrepeated = false; 5 for (var j = 0, Len = result.length; J The general idea is to move the elements of the array to another array one by one, checking to see if the element is duplicated in the process of handling it, and throwing it away if any. As y

Referenced from: http://www.cnblogs.com/sosoft/archive/2013/12/08/3463830.htmlThe array type does not provide a way to repeat, and if you want to kill the repeating elements of the array, you have to do it yourself:1functionUnique (arr) {2var result =[], isrepeated;3for (var i = 0, len = arr.length; i ) {4 isrepeated =False;5for (var j = 0, len = result.length; J ) {6if (arr[i] = =Result[j]) {true;break; 9 }10 }11 if (! isrepeated) {12 Result.push (Arr[i]); 13 }14 }15 return Result;