Given an array of integers, rotate it one position.

Question

Doug · Accepted Answer

1-divide the array in 2 of rotation order (e.g. num rotation =3 , then a1[N-3] , a2[3-N]
2-Reverse the first array
3-reverse the second array
3-reverse the whole array

Note that the functions that do the reverse is not creating new arrays. Then with this you have time complexity O(n) and space complexity O(1) because you're using only one array in memory

Anonymous · Answer

I used a queue to record the array, pop and then push.

Zach · Answer

The general case (rotate k positions)

int[] rotateArray(int[] A, int k)
{
if (k==0)
 return A; // rotating 0 positions is the original array

int savedVal = A[k-1%A.length]
for (int i = 0; i < A.length; ++i)
{
  int temp = A[(i+k)%A.length];
  A[(i+k)%A.length] = savedVal;
  savedVal = temp;
}
A[k-1%A.length] =  savedVal;

return A;
}

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