Problem
You are given the head of a linked list, which contains a series of integers separated by 0’s. The beginning and end of the linked list will have Node.val == 0.
For every two consecutive 0’s, merge all the nodes lying in between them into a single node whose value is the sum of all the merged nodes. The modified list should not contain any 0’s.
Return the head of the modified linked list.
Testcases
Input: head = [0,3,1,0,4,5,2,0]
Output: [4,11]
Initial Result
Time spent: –
Fail. I had the same pointer design but incorrectly tracked dummy.
Initial Approach
I had a dummy pointer that would sit at the first zero in our two-zero sequence and then used a second pointer for the second zero accordingly. This was flawed since the problem actually wants us to remove the zeroes bounding each range to summate, so I modified the arrangement to keep dummy right before a dedicated pointer for the first zero. Then, I used an integer to summate the values in between and stored the results in a new node between the dummy and the node following the second zero.
What I Missed
I was adjusting dummy and then returning the same pointer, leading to an incomplete list. I also failed to disconnect the final zero (we expect to end up with one zero at the end stored at first_0).
Approach to the Solution
Perform the same logic as above with disconnecting of dummy and an early return whenever there is one zero and no second zero to accompany it.
Key Takeaways
Draw out pointers for linked list problems.