LeetCode 2. Add Two Numbers [Medium] [C++] 解題筆記

這題給我們兩個 Linked List，並保證最後一個 node 值都不為 0，要求我們將兩個 list 的reverse order 當成兩個數字相加。

EX:

         Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

        Output: 7 -> 0 -> 8 ,  342 + 465 = 807

想法:

    這題就是當純的實作題，沒啥特別技巧，注意進位和 head 的特殊情況即可。

Time Complexity: O(max(m,n)), m,n is len of l1, l2,receptively.

完整程式碼:

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode *head = new ListNode(0);
        ListNode *cur = head;
        int carry = 0;
        
        while (l1 != nullptr && l2 != nullptr) {
            auto sum = l1->val + l2->val + carry;
            cur->next = new ListNode(sum % 10);
            carry = sum / 10;
            cur = cur->next;
            l1 = l1->next;
            l2 = l2->next;
        }
        
        while (l1 != nullptr) {
            auto sum = l1->val + carry;
            cur->next = new ListNode(sum % 10);
            carry = sum / 10;
            l1 = l1->next;
            cur = cur->next;
        }
        
        while (l2 != nullptr) {
            auto sum = l2->val + carry;
            cur->next = new ListNode(sum % 10);
            carry = sum / 10;
            l2 = l2->next;
            cur = cur->next;
        }
        if (carry > 0) {
            cur->next = new ListNode(carry);
            carry = 0;
            cur = cur->next;
        }
        
        return head->next;
    }
};