Create a dummy head node for the result list. A dummy node simplifies linked list construction because we always append new nodes after it. This avoids special handling for the first node. 2. Initialize a pointer called current. This pointer tracks the last node in the result list so new digits can be appended efficiently. 3. Initialize carry = 0. Carry stores overflow from the previous digit addition. 4. Traverse both linked lists simultaneously. Continue while at least one list still has nodes or a carry still exists. 5. Extract the current digit values. If a list still has nodes, use its digit value. Otherwise use 0. This handles lists of unequal length naturally. 6. Compute the digit sum.
Link: https://brain.tamnd.com/practice/leetcode/00xx/2/
Create a dummy head node for the result list. A dummy node simplifies linked list construction because we always append new nodes after it. This avoids special handling for the first node. 2. Initialize a pointer called current. This pointer tracks the last node in the result list so new digits can be appended efficiently. 3. Initialize carry = 0. Carry stores overflow from the previous digit addition. 4. Traverse both linked lists simultaneously. Continue while at least one list still has nodes or a carry still exists. 5. Extract the current digit values. If a list still has nodes, use its digit value. Otherwise use 0. This handles lists of unequal length naturally. 6. Compute the digit sum.
Link: https://brain.tamnd.com/practice/leetcode/00xx/2/