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.
Was browsing through Medium coding problems of Leetcode( Oh yeah. I think Im at this level ) when I stumble upon this one — validating binary tree. Pfft. You dummy, NO WAY this is medium. Turns out it is harder than I thought ^^, not because it is hard but because of my stupid mistakes….
It’s actually not that hard…..It’s just …… verbose and time consuming. Well, let’s dive in, shall we.
Times and times again, here we are for the new exciting yet challenging semester. This time I am taking AI. I set my mind up for the challenge, to redress some of the problem I had encountered last semester.