A variant of the Fibonacci sequence on practical issues

We can use the small rectangle of 2*1 to cover the larger rectangle horizontally or vertically. Can you tell me how many methods 1 use the array structure traversal method if the small rectangle with n 2*1 covers a 2*n large rectangle without overlapping (target==1 | |  target==0)             return 1;         int [] arr = new int [target+1];             arr[0] = 1;             arr[1] = 1;         for (int i=2;i<=target;i++) {             arr[i] = arr[i-1]+arr[i-2];        }  &NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;RETURN&NBSP;ARR[TARGET];2 uses two int variables and a temporary variable in a while statement without the help of the array structure temp    If (target==1 | |  target==0)             return 1;         int num1 = 1,num2 = 1;         while (target>1) {             Int temp = num2;            num2  +=num1;            num1=temp;             target--;         }        return num2;    3 recursive efficiency minimum         if (target==1 | |  target==0)             return 1;         if (target>1)              return rectcover (target-1) +rectcover (target-2);         return 0; here because of the use of OJ, when I above the   if (target==1 | |  target==0)             return 1; wrote a   if (target==1)             return 1;   if (target==0)             return 1;  if (target>1)             return  Rectcover (target-1) +rectcover (target-2);         return 0;o J System on the judgment time has timed out and really think about the problem that can be judged once, why do you want two times?

A variant of the Fibonacci sequence on a practical issue