C language realizes the creation of the list, calculates the chain list length and the merged __c language of two lists

problem Description: Input two series from keyboard, form two single linked list

(1). Calculates the length of two single linked lists (2). Output the maximum, minimum, and average (3) of a long list. Counts two of the same elements in the list (4). Merge two linked lists and output

Program code:

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>

struct MyList
{
	int data;
	struct MyList *next;

/*......................................
	The tail inserts the method to create the chain list ... .... * * * struct mylist *createlist (void) {int i=0;; * * * * * * * * * * * * *;
	int x;
	struct MyList *s,*q,*h;
	h= (struct mylist *) malloc (sizeof (struct mylist));
	h->next=null;
	Q=h;
	printf ("Enter the value of the%d node (enter 0 end): \ n", i+1);
	scanf ("%d", &x);
		while (x!=0) {i=i+1;
		s= (struct mylist *) malloc (sizeof (struct mylist));
		s->data=x;
		q->next=s;
		printf ("Enter the value of the%d node (enter 0 end): \ n", i+1);
		scanf ("%d", &x);
		s->next=null;
	S=1;
	printf ("\ n");
return (h); }

/*...................................
	Find the list of the table long ... ..... * * * int getlength (struct mylist *head) {int i=0;;;.. * * * * * * int.) (+).
	struct MyList *p;
	if (Head->next==null) return 0;
		else {p=head->next;
			while (p) {i++;
		p=p->next;
	return i; }
}

/*.....................
	To find the maximum, minimum, and average value of all elements in a list ... * * * void Getmna (struct mylist *head1) {int max,min;
	float average;
	int sum=0,n=0;
	int i;
	int a[100];
	if (head1->next==null){printf ("Linked list is empty list!\n");
	Exit (0);
		while (head1->next!=null) {a[n]=head1->next->data;
		n++;
	head1=head1->next;
	} Max=a[0];
	MIN=A[0];
	SUM=A[0];
		for (i=1;i<n;i++) {sum=sum+a[i];
		if (A[i]>max) max=a[i];
	if (a[i]<min) min=a[i];
	} average=sum/(n*1.0);
	printf ("The maximum, minimum and average value of the list is:%d,%d,%.3f\n", max,min,average);
printf ("\ n"); }

/*.........................
	Count the number of the same elements in two linked lists .... * * * * void Stalink (struct mylist *head2,struct mylist *head3) {int count=0,i=0,j=0);
	int m,n;
	int arr1[100];
	int arr2[100];
		while (head2->next!=null) {arr1[i]=head2->next->data;
		i++;
	head2=head2->next;
	} m=i;
		while (head3->next!=null) {arr2[j]=head3->next->data;
		j + +;
	head3=head3->next;
	} n=j;
		for (i=0;i<m;i++) {for (j=0;j<n;j++) {if (arr1[i]==arr2[j]) count=count+1;
	The number of the same elements in two linked lists is:%d \ n, count.
printf ("\ n"); }

/*.................................. Merge and export two linked lists ........... ......... */void Addlink (struct mylist *head4,struct mylist *head5) {struct mylist;;;;;
	PR=HEAD4;
	while (Head4->next!=null) head4=head4->next;
	head4->next=head5->next;
	printf ("Two merged list: \ n");
		while (Pr->next!=null) {printf ("%d", pr->next->data);
	pr=pr->next;
printf ("\ n \ nthe"); }

/*.....................................
	The main program ....... */int main (void) {int length1;;..... * * * (=) (=)--------.
	int length2;
	struct MyList *my1;
	struct MyList *my2;
	printf ("Enter first sequence: \ n");
	My1=createlist ();
	printf ("Enter second sequence: \ n");
	My2=createlist ();
	Length1=getlength (MY1);
	Length2=getlength (MY2);
	printf ("First linked list and second linked list are%d and%d\n in length, respectively", length1,length2);
	printf ("\ n");
		if (length1>=length2) {printf ("the maximum, minimum, and average value of the first linked list: \ n");
	Getmna (MY1);
		else {printf ("the maximum, minimum, and average value of the second list: \ n");
	Getmna (MY2);
	printf ("Now compares the number of identical elements in two linked lists: \ n");
	Stalink (MY1,MY2);
	printf ("Two linked lists are merged: \ n");
	Addlink (MY1,MY2);
return 0; }

Program Run Result: