remove duplicates from ipod

title :Given a sorted array, remove the duplicates in place such, all element appear only once and return the new L Ength.Do the allocate extra space for another array, and you must does this on place with constant memory.For example,Given input Array A = [1,1,2] ,Your function should return length = 2 , and A is now [1,2] .code : OJ Test via runtime:143 ms1 classSolution:2 #@param a list of integers3

Given a sorted array, remove the duplicates in place such, all element appear only once and return the new L Ength.Do the allocate extra space for another array, and you must does this on place with constant memory.For example,Given input array nums = [1,1,2] ,Your function should return length = 2 , with the first of the elements of nums being and 1 2 Respectivel Y. It doesn ' t matter what are you leave b

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;

JS array type does not provide a way to repeat, if you want to kill the array of repeating elements, you can expand it yourself.The first idea is to sort the array first, and then compare the elements before and after they are equal, or continue, otherwise they will be recorded in the return value:Array.prototype.unique =function () { vartemp = []; This. sort (); varLen = This. Length; for(i = 0; i ) { if( This[I] = = This[I+1]) { Continue; } Temp[temp.length]= This[i];

82. Remove Duplicates from Sorted List II, duplicatessorted Given a sorted linked list, delete all nodes that have duplicate numbers, leaving onlyDistinctNumbers from the original list. For example,Given1->2->3->3->4->4->5, Return1->2->5.Given1->1->1->2->3, Return2->3. 1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) :

Because the test instructions of this topic is to require us to operate on the original array, the operation becomes slightly more complicated, otherwise the direct use of map is the simplest.The basic idea is to record two pointers, one is the current array, the other is the destination array, and note that if more than 2 repetitions are found, then the cur of the destination array will be blocked,Until a different occurrence and then assign a value forward.Class Solution {public: int remove

/** 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. Remove

Removes a repeating element from the linked list, given a sorted list.Ideas:Because they are already sorted, it is possible to determine whether or not the adjacent elements are equal if they are duplicated.1 traversal, using the new linked list TMP to save the non-repeating elements of the linked list head, return to the new linked list.Implementation code:public class Solution {public ListNode deleteduplicates (ListNode head) { if (head = = NULL | | head.next = = NULL) { return he

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. Think more about it. Train yourself for more tra

Total accepted:59433 Total submissions:230628 difficulty:medium 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. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode* Deleteduplicates (listnode*head) {ListNode* dele

Given a sorted linked list, delete all nodes that has duplicate numbers, leaving only distinct numbers from the Original list.ExampleGiven 1->2->3->3->4->4->5 , return 1->2->5 .Given 1->1->1->2->3 , return 2->3 ./*** Definition for ListNode * public class ListNode {* int val; * ListNode Next; * ListNode (int x) {* val = x; * next = NULL; * } * } */ Public classSolution {/** * @paramListNode Head is the head of the linked list *@return: ListNode head of the linked list*/ Public Static

/*** Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int X) {val = x;}}*/ Public classSolution { PublicListNode deleteduplicates (ListNode head) {MapNewHashmap(); if(head==NULL) return NULL; ListNode Temp=Head; ListNode Tail=NULL; while(temp!=NULL) { if(Mp.containskey (temp.val)) {if(tail!=NULL) Tail.next=Temp.next; ElseHead=Temp.next; } Else{mp.put (Temp.val,1); Tail=temp; } temp=Temp.next; } ret

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), Next (NULL) {}7 * };8 */9 classSolution {Ten Public: Onelistnode* Deleteduplicates (listn

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* Deleteduplicates (listnode*head) { A if(!head) { - returnhead; -

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; * ListNode *next; * ListNode (int x): Val (x), Next (NULL) {} *}; */class Solution {public:l

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 .Test instructionsGive a list and delete all duplicate nodes.Steps:(1) Define a new head node that points to the head and moves the head forward one bit.1, 1, 1, 2, 3To head, 1, 1, 1, 2, 3(2) Traverse each node, compare two nodes at a time, Head.next and Head.next.next.1. If Head.nex

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 deleteduplicates (ListNode head) { One if(Head = =NULL|| Head.next = =NULL) { A

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){ if(previous!=NULL){ if(dp.val==previous.val) {Previous.next=Dp.next; }

/** * Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x): Val (x), Next (NULL) {} *}; */class Solution {//on_my_ownpublic:listnode* deleteduplicates (listnode* head) {listnode *ret=new ListNode (0); The final returned pointer ListNode *ret1=ret;//represents the RET to be linked to the node's pointer listnode *pre=head;//the first node of each repeating group ListNode *cur=hea d;//the last node of each repeating group bool FIRST=TRUE;//1 represen