Today Leetcode problem is rotating linked list . Doing it is a no brainer( Answer below). What I suddenly realize is you could pipe the boolean expression to simplify the boolean check later on. To jump straight on:
int k1 = (l == 0 ) ? 0 : k % l; /* equivalent to if(l != 0 ) k1 = k % l; else k1 = 0; */ if(k1 == 0) return head; //rotate with k multiple of list length or list empty
It is like a filter so the end result is just one simple check. I know it won’t make much of a difference but it comes to my mind naturally.