Move a string of numbers backward in sequence

                                                            Move a series of numbers backward

One 、 Title Description

　　 Yes n It's an integer , Move the front numbers back in order m A place , Last m The number becomes the first m Number （m<n）.( More , Please see the Programmers are on the road )

Two 、 Analysis answers

　　 This problem is mainly about the training of programming logic . It's about arrays 、 Knowledge points of pointer , This topic can also well express the choice of different data structures , There will be completely different ideas for solving the same problem , Programming complexity will also vary greatly .
　　 Method 1 ： use Array Data structures store data . Ideas ：① Use a temporary array , Put the last... In the original array m First copy the numbers into the new array , Then move the elements in the original array backward m A place , Finally, copy the elements in the temporary array to the beginning of the original array .② Use a temporary array with the same length as the original array , After the original array m Numbers are copied to the front of the temporary array m position , Then copy the remaining numbers of the original array to the back of the temporary array , Then return the temporary array .
　　 It is worth noting that the idea ①② The size of the temporary array space is different , There is a slight difference in programming . Also pay attention to the calculation of array subscripts , This is easy to make mistakes .

#include<stdio.h>
void shift(int *a,int n,int m)
{
    int t[20];
	int i;
	for(i=0;i<n;i++)
	{
		t[i]=a[i];
	}
	for(i=0;i<m;i++)
	{
		a[i]=t[n-m+i];
	}
	for(i=m;i<n;i++)
	{
		a[i]=t[i-m];
	}	
}
int main()
{
	int a[20];
	int n,m;
	int i;
	
	scanf("%d",&n);

	for(i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
		}

	    scanf("%d",&m);
		shift(a,n,m);
		for(i=0;i<n;i++)
		{
			printf("%d \t ",a[i]);
		}
		printf("\n");


		return 0;
}

　　 Method 2 ： use Single chain list Data structures store data . Ideas ： Will be the first （N - M） The element of the position is the first node （ That's the end m The first element node of an element ）, The first node element of the original linked list can be linked to the end node of the original linked list .

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

typedef struct data_node{

	int data;

	struct data_node *next;

}Node;

int main(){

	Node *head = NULL , *tail =NULL,  *p = NULL;

	int n,m,i=0;

	// Input N It's an integer 
	scanf("%d",&n);

	while(i++<n){
	
		// The memory space occupied by the application node 
		if( (p = (Node *)malloc(sizeof(Node))) == NULL){
		
			printf("memery is not available");
			exit(1);

		}

		scanf("%d", &(p->data));
		p->next = NULL;

		// The current application is for the first node 
		if(head == NULL){

			head = tail = p;
	
		}else{
			// Tail interpolation 
		
			tail->next = p;
			tail = p;
		
		}	
	}


	// Find No (N-M) Location of the precursor node .
	scanf("%d",&m);

    i=0;
	p = head;
    while( i++<n -m -1){
	
      p = p->next;

	}

	// Notice the exchange of these elements . Is the core of realizing this topic 
	// Last m The first element of the elements acts as the chain head , The first node element of the original linked list is linked to the end node of the original linked list . In this way, the backward shift is realized 
	tail->next = head;
	head = p->next;
	
	tail = p;
	tail->next = NULL;

	// Output processed number 
	p = head;
	while(p){
	
		printf("%d  ",p->data);

		p = p->next;
	}
	printf("\n");

	return 0;
}

　　 Schematic diagram of program running results ：

　　

3、 ... and 、 Summary

　　 From the above two different methods to solve the same problem, we can see , Choice of different data structures , It has a great impact on programming ideas , The ingenious choice of data structure sometimes greatly simplifies the understanding of programming logic . meanwhile , Different data structures will also make a great difference in the time complexity and space complexity of the program algorithm .