置換元素和非置換元素

Problem statement: Write a c program to cyclically permute the element of an array. (In right to left direction). Array should be taken as input from the user.

問題陳述：編寫一個c程序來循環置換array的元素 。 (從右到左方向)。 數組應作為用戶的輸入。

Explanation with example:

舉例說明：

Let the user input for the array be: 4 5 6 7 8 10 11 34 56 1

讓用戶輸入該數組為： 4 5 6 7 8 10 11 34 56 1

The cyclic permutation operation on the array results in rotation of the array by one position in right to left direction.

陣列上的循環置換操作導致陣列從右到左方向旋轉一個位置。

Thus the array becomes: 5 6 7 8 10 11 34 56 1 4

因此，該數組變為： 5 6 7 8 10 11 34 56 1 4

i.e. the first element becomes the last element & the rest of the elements are shifted by one position.

也就是說，第一個元素變為最后一個元素，其余元素移位一個位置。

Algorithm:

算法：

    To shift the (i+1)th element to the left 
can be easily done only by:
A[i] =A[i+1]; // A is the input array
So it may seem that the entire shifting can be done by
For i =0:n-1
A[i] =A[(i+1)%n];
End For
But this will lead to wrong solution since for i=n-1
A[n-1]=A[0] which is correct statement but A[0] has been 
updated already. This code snippet will result in 
A[0] & A[n-1] to be same which is actually wrong. 
So what we need to do is to store A[0] ( staring element) 
and assign this value to A[n-1]
Thus, a little modification can lead to correct solution:
1.  Set temp to A[0]
2.  For i=0:n-1
If i==n-1
A[i]=temp; //actually it means A[n-1]=A[0]
Else
A[i]=A[i+1];
End If
End For
3.  Print the updated array.

循環置換數組元素的C實現 (C Implementation for Cyclically Permute the Elements of an Array)

#include <stdio.h>
#include <stdlib.h>
//function to print the array
void print(int* a,int n){
printf("printing ........\n");
for(int i=0;i<n;i++)
printf("%d ",a[i]);
printf("\n");
}
int* cyclicallyPermute(int* a,int n){
int temp=a[0];//store a[0]
for(int i=0;i<n;i++){
//for the last element in the modified array 
//it will be starting elemnt    
if(i==n-1) 
a[i]=temp;
//for other element shift left    
else
a[i]=a[i+1];
}
return a;    
}
int main()
{   
int n;
printf("enter array length,n: ");
scanf("%d",&n);
//allocating array dynamically
int* a=(int*)(malloc(sizeof(int)*n));
printf("enter elements: \n");
//taking input
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
printf("array before permutation\n");
print(a,n);
//function to permute cyclically
//returning base adress of modified array
a=cyclicallyPermute(a,n);
printf("array after permutation\n");
print(a,n);
return 0;
}

Output

輸出量

enter array length,n: 10
enter elements:
4 5 6 7 8 10 11 34 56 1
array before permutation
printing ........
4 5 6 7 8 10 11 34 56 1
array after permutation
printing ........
5 6 7 8 10 11 34 56 1 4 

翻譯自: https://www.includehelp.com/c-programs/cyclically-permute-the-elements-of-an-array.aspx

置換元素和非置換元素