2. Add Two Numbers (inplace... but wrong)

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode * reverseList(ListNode* head){ if (head == nullptr || head->next == nullptr) return head; ListNode * newhead = reverseList(head->next); ListNode * curr = newhead; while(curr->next != nullptr) curr = curr->next; // find the last node of the new list curr->next = head; head->next = nullptr; return newhead; } ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { if (l1 == nullptr) return l2; else if (l2 == nullptr) return l1; ListNode * newl1 = reverseList(l1); ListNode * newl2 = reverseList(l2); ListNode * l1ptr = newl1; // store results on l1ptr ListNode * l2ptr = newl2; int carry = 0; // store carry while(l1ptr->next != nullptr && l2ptr->next != nullptr){ if (l1ptr->val + l2ptr->val + carry <= 9){ l1ptr->val += l2ptr->val + carry; carry = 0; } else{ l1ptr->val = (l1ptr->val + l2ptr->val + carry) - 10; carry = 1; } l1ptr = l1ptr->next; l2ptr = l2ptr->next; } // exited while loop. Now we are dealing with the last node of one of the lists (or both) if (l1ptr->val + l2ptr->val + carry <= 9){ l1ptr->val += l2ptr->val + carry; carry = 0; } else{ l1ptr->val = (l1ptr->val + l2ptr->val + carry) - 10; carry = 1; } if (l1->next == nullptr) // redirect the last node of newl1 to points to remaining nodes of newl2 l1ptr->next = l2ptr->next; // start working on the remaining nodes l1ptr = l1ptr->next; while(l1ptr != nullptr){ if (l1ptr->val + carry <= 9){ l1ptr->val += carry; carry = 0; } else{ l1ptr->val = (l1ptr->val + carry) - 10; carry = 1; } l1ptr = l1ptr->next; } // finished. Nodes stored in the list whose head is newl1; return reverseList(newl1); } };
- Reverse list : note how to find the second to last node!! Use curr->next != nullptr, not curr != nullptr... Or else will find the last node instead...

The algo is :
- First reverse both lists.

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