The task is to rotate left or rotate right a subarray of an array given number of times.
Let me explain this on an example:
There's a trick to this. It's pretty weird that you'd get this for homework if the trick wasn't mentioned in class. Anyway...
To rotate a sequence of N elements left by M:
done
e.g. left by 2: 1234567 -> 7654321 -> 7654312 -> 3456712
Here is my code, it makes exactly n reads and n writes, where n is subarray size.
#include<iostream>
int arr[]= { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
// replacing 'addr( pos, from, size )' with just 'pos' mean rotation the whole array
int addr( int ptr, int from, int size)
{
return (ptr + from ) % size;
}
void rotate( int* arr, int shift, int from, int count, int size)
{
int i;
int pos= 0;
int cycle= 0;
int c= 0;
int c_old= 0;
// exactly count steps
for ( i=0; i< count; i++ ){
// rotation of arrays part is essentially a permutation.
// every permutation can be decomposed on cycles
// here cycle processing begins
c= arr[ addr( pos, from, size ) ];
while (1){
// one step inside the cycle
pos= (pos + shift) % count;
if ( pos == cycle )
break;
c_old= c;
c= arr[ addr( pos, from, size ) ];
arr[ addr( pos, from, size ) ]= c_old;
i++;
}
// here cycle processing ends
arr[ addr( pos, from, size ) ]= c;
pos= (pos + 1) % count;
cycle= (cycle + 1) % count;
}
}
int main()
{
rotate( arr, 4, 6, 6, 11 );
int i;
for ( i=0; i<11; i++){
std::cout << arr[i] << " ";
}
std::cout << std::endl;
return 0;
}