c语言i++和++i程序_使用C ++程序从链接列表中消除重复项

c语言i++和++i程序

Given a sorted linked list (elements are sorted in ascending order). Eliminate duplicates from the given LL, such that output LL contains only unique elements.

给定一个排序的链表(元素按升序排序)。从给定的LL中消除重复项，以便输出LL仅包含唯一元素。

Input format: Linked list elements (separated by space and terminated by 1)

输入格式：链接列表元素(以空格分隔并以1终止)

    Sample Input 1 :
1 2 3 3 3 4 4 5 5 5 7 -1
Sample Output 1 :
1 2 3 4 5 7

Description:

描述：

In this question, we are given a sorted linked list with duplicate elements in it. Our task is to remove the duplicate nodes of the linked list. Since the list is sorted, the duplicate elements will be present in consecutive orders, that makes the work a lot easier.

在这个问题中，我们得到一个包含重复元素的排序链表。我们的任务是删除链表的重复节点。由于列表已排序，因此重复的元素将以连续的顺序显示，这使工作变得更加容易。

Example:

例：

    Lets the list be:
1->2->3->3->4->4->4->NULL
The modified list will be:
1->2->3->4->NULL

Solution Explanation:

解决方案说明：

To solve this problem, we keep the temp pointer pointing to node. While(temp -> next != NULL), we check if the data of temp is equal to data of temp->next. If they are equal, we will point the temp-> next to temp -> next ->next.
Original LL: 3-> 3-> 4-> ...
After Change: 3-> 4-> ... Leaving out the middle 3->

为了解决这个问题，我们使临时指针指向节点。 while(temp-> next！= NULL)时，我们检查temp的数据是否等于temp-> next的数据。如果它们相等，则将temp->指向temp-> next-> next。
原始LL：3-> 3-> 4-> ...
更改后：3-> 4-> ...省略中间3->

This will keep on until we reached the end of the list and return head at the end.

这将一直持续到我们到达列表的末尾并返回末尾。

Steps

脚步

    1.	1->2->3->3->4->4->4->NULL, temp = 1
2.	1->2->3->3->4->4->4->NULL, temp = 2
3.	1->2->3->3->4->4->4->NULL, temp = 3
4.	1->2->3->4->4->4->NULL, temp = 4
5.	1->2->3->4->4->NULL, temp = 4
Finally,
1->2->3->4->NULL

Function:

功能：

Node *removeDuplicate(Node* head){
//temp pointing to head
Node *temp = head;            
while(temp->next != NULL && temp != NULL){
//Duplicate Found
if(temp->data == temp->next->data){      
//DUplicate Removed
temp -> next = temp ->next ->next;    
}
else{
//No Duplicate Present
temp = temp->next;               
}
}
//Return Head
return head;                
}

C++ Code:

C ++代码：

#include<bits/stdc++.h>
using namespace std;
struct Node{// linked list Node
int data;
Node * next;
};
Node *newNode(int k){ //defining new node
Node *temp = (Node*)malloc(sizeof(Node)); 
temp->data = k; 
temp->next = NULL; 
return temp; 
}
//Used to add new node at the end of the list
Node *addNode(Node* head, int k){
if(head == NULL){
head = newNode(k);
}
else{
Node * temp = head;
Node * node = newNode(k);
while(temp->next!= NULL){
temp = temp->next;
}
temp-> next = node;
}
return head;
}
// Used to create new linked list and return head
Node *createNewLL(){
int cont = 1;
int data;
Node* head = NULL;
while(cont){
cout<<"Enter the data of the Node"<<endl;
cin>>data;
head = addNode(head,data);
cout<<"Do you want to continue?(0/1)"<<endl;
cin>>cont;
}
return head;
}
//To print the Linked List
void *printLL(Node * head){
while(head!= NULL){
cout<<head->data<<"->";
head = head-> next;
}
cout<<"NULL"<<endl;
}
//Function 
Node *removeDuplicate(Node* head){
//temp pointing to head
Node *temp = head;
while(temp->next != NULL && temp != NULL){
//Duplicate Found
if(temp->data == temp->next->data){
//DUplicate Removed
temp -> next = temp ->next ->next;
}
else{
//No Duplicate Present
temp = temp->next;
}
}
//Return Head
return head;
}
//Driver Main
int main(){
Node * head = createNewLL();
cout<<"The linked list is"<<endl;
printLL(head);
head = removeDuplicate(head);
cout<<"The new Linked List is" <<endl;
printLL(head);
return 0;
}

Output

输出量

Enter the data of the Node
1
Do you want to continue?(0/1)
1
Enter the data of the Node
1
Do you want to continue?(0/1)
1
Enter the data of the Node
2
Do you want to continue?(0/1)
1
Enter the data of the Node
2
Do you want to continue?(0/1)
1
Enter the data of the Node
3
Do you want to continue?(0/1)
1
Enter the data of the Node
4
Do you want to continue?(0/1)
1
Enter the data of the Node
5
Do you want to continue?(0/1)
1
Enter the data of the Node
5
Do you want to continue?(0/1)
1
Enter the data of the Node
5
Do you want to continue?(0/1)
0
The linked list is
1->1->2->2->3->4->5->5->5->NULL
The new Linked List is
1->2->3->4->5->NULL