将数组的元素添加到链表中

Adding elements of array into linked list

本文关键字：链表 添加 元素 数组      更新时间：2023-10-16

我有数组元素ar1[i]我想将此数组的数据添加到链表中

struct node
{
int data;
struct node*next;
};

void create(struct node *head,int ar1[] int i1)
{
struct node *temp, *p, *q;
int i;

head = (struct node*)malloc(sizeof(struct node));
temp = (struct node*)malloc(sizeof(struct node));

for (i=0; i<i1; i++) //lets say i1 is no. of data in array
{
if(head == NULL)
{
head->data = ar1[i];
head->next = NULL;
}
else
{
temp->data = ar1[i];
temp->next = NULL;
p = head;
while(p->next != NULL)
p = p->next;
p->next = temp;
}
}

我做了这个程序，但它不起作用。它接受输入数据但不显示输出，也不终止程序。

编辑：正如大家在评论中建议的那样，我尽力了并提出了解决方案。实际上，我的程序是从用户那里获取最多 10 位数字并对其执行算术运算。

这是我的代码：

/*
WAP to store at most 10 digit integer in a Singly linked list and 
perform
arithmetic operations on it.
*/
#include<iostream>
using namespace std;
struct node
{
int data;
struct node*next;
};
void create(struct node *head,int ar1[],int ar2[], int i1, int i2)
{
struct node *newNode, *temp;
int i=0;
head = (struct node *)malloc(sizeof(struct node));
if(head == NULL)
{
cout<<"Unable to allocate memory.";
}
else
{
head->data = ar1[i]; // Link the data field with data
head->next = NULL; // Link the address field to NULL
temp = head;
for(i=1; i<i1; i++)     //Create n nodes and adds to linked list
{
newNode = (struct node *)malloc(sizeof(struct node));
if(newNode == NULL)     
{
cout<<"Unable to allocate memory.";
break;
}
else
{

newNode->data = ar1[i]; 
newNode->next = NULL;
temp->next = newNode; 
temp = temp->next; 
}
}
}
temp = head;
while(temp!=NULL)
{
cout<<temp->data<<" ";
temp= temp->next;
}   
}
int main()
{
struct node*head = NULL ;
int n1,n2;
int i1,i2,choice;
int ar1[10],ar2[10];
head = (struct node*)malloc(sizeof(struct node));
cout<<"Enter numbers you want to perform operations onn1). ";
cin>>n1;
cout<<"2). ";
cin>>n2;
i1=0; i2=0;
while(n1)       //getting each digit of given number
{
ar1[i1] = n1 %10;
n1 /= 10;
i1= i1 + 1;     
}
while(n2)       //getting each digit of given number
{
ar2[i2] = n2 % 10;
n2 /= 10;
i2++;   
}
cout<<"nChoose what operation you want to 
perform:n1.Additionn2Subtractionn";
cin>>choice;
create(head,ar1,ar2,i1,i2);
/*if(choice == 1)  I comment out this part i.e. not important rn.
{
add(ar1,ar2);
}
else if (choice == 2)
sub();
else
{
cout<<"Please chose valid datan";
}
*/
return 0;
}

有很多问题。create函数的整个逻辑是错误的：

void create(struct node *head, int ar1[], int i1)
{
struct node *temp, *p, *q;
int i;
head = (struct node*)malloc(sizeof(struct node));  // Problem 1
temp = (struct node*)malloc(sizeof(struct node));
for (i = 0; i < i1; i++)
{
if (head == NULL)                                // Problem 2
{
head->data = ar1[i];  
head->next = NULL;
}
else
{
temp->data = ar1[i];
temp->next = NULL;
p = head;
while (p->next != NULL)
p = p->next;
p->next = temp;                                // Problem 3
}
}
}

...
create(myhead, myarray, maysize);                    // Problem 4
...

问题1

你分配内存太早了，你应该只在你真正需要的时候分配它。

问题2

部分由问题 1 引起：head不能在这里NULL(假设malloc没有失败，但那是另一回事(，因为您之前已将其分配给非NULL值。

问题3

由于上面的while循环，这里总是p->nextNULL。

问题4

假设你不能create喜欢这个：

...
myhead = NULL;
...
create(myhead, myarray, maysize);

myhead在调用create后仍将NULL，您应该阅读这篇SO文章。

只需使用纸和铅笔来理解逻辑，然后 按顺序优化代码。

struct node
{
int val;
struct node *next;
};
struct node *head=NULL, *last=NULL;
void insert( int value)
{
struct node *tmp;
tmp=(struct node *)malloc(sizeof(struct node));
tmp->val=value;
tmp->next=NULL;
if( head == NULL )
{
head=tmp;
last=tmp;
}
else
{
last->next=tmp;
last=tmp;
}
}
void printlist()
{
struct node *tmp=head;
printf("The list is:");
while( tmp != NULL )
{
printf(" %d", tmp->val);
tmp=tmp->next;
}
printf("n");
}
void search( )
{
printf("Which value to Search?");
int n, k=0;
scanf("%d",&n);
struct node *tmp=head;
while( tmp != NULL )
{
if( tmp->val == n )
k=1;
tmp=tmp->next;
}
if( k )
printf("Yesn");
else
printf("Non");
}
void deletenode( )
{
int n, posi;
printf("Where to Delete?n");
printf("1tFirst Positionn");
printf("2tLast Positionn");
printf("3tAny Desired Positionn");
scanf("%d",&n);
if( n == 1 )
{
struct node *tmp=head;
tmp=tmp->next;
head=tmp;
printlist();
}
else if( n == 2 )
{
struct node *tmp=head ;
while( 1 )
{
if( tmp->next->next == NULL )
{
tmp->next=NULL;
break;
}
else
tmp=tmp->next;
}
printlist();
}
else if( n == 3 )
{
int i=1;
printf("Enter Position: ");
scanf("%d",&posi);
struct node *tmp=head;
while( tmp->next != NULL && i <= posi )
{
if( i+1 == posi )
{
tmp->next=tmp->next->next;
break;
}
else
{
i++;
tmp=tmp->next;
}
}
printlist();
}
}
void ins()
{
int n, v, posi;
printf("Where to insert?n");
printf("1tFirst Positionn");
printf("2tLast Positionn");
printf("3tAny Desired Positionn");
scanf("%d",&n);
if( n == 1 )
{
printf("Enter value: ");
scanf("%d",&v);
struct node *tmp;
tmp=(struct node *)malloc( sizeof(struct node));
tmp->val=v;
tmp->next=head;
head=tmp;
printlist();
}
else if( n == 2 )
{
printf("Enter value: ");
scanf("%d",&v);
struct node *tmp;
tmp=(struct node *)malloc(sizeof(struct node));
tmp->val=v;
tmp->next=NULL;
last->next=tmp;
last=tmp;
printlist();
}
else if(n == 3 )
{
int i=1;
printf("Enter position: ");
scanf("%d",&posi);
printf("nEnter value: ");
scanf("%d",&n);
struct node *tmp1, *tmp=head, *tmp2;
tmp1=(struct node *)malloc(sizeof(struct node));
tmp1->val=n;
tmp1->next=NULL;
while( tmp->next != NULL && i <= posi)
{
if( i+1 == posi )
{
tmp2=tmp->next;
tmp->next=tmp1;
tmp1->next=tmp2;
}
i++;
tmp=tmp->next;
}
printlist();
}
}
int main()
{
int n;
printf("Enter value & Enter -1 to exit.n");
while( 1 )
{
scanf("%d",&n);
if( n < 0 )
break;
insert(n);
}
printlist();
printf("Enter choice:nn");
printf("1tInsertn");
printf("2tDeleten");
printf("3tSearchn");
printf("4tExitn");
scanf("%d",&n);
if( n == 1 )
ins();
else if( n == 2 )
deletenode();
else if( n == 3 )
search();
else if( n == 4 )
return 0;
return 0;
}
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