2. Add Two Numbers

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My Solution

The Key Idea for Solving This Coding Question

C++ Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode *dummy = new ListNode(), *curr = dummy;
        int carry = 0;

        while (l1 != nullptr || l2 != nullptr) {
            int sum = carry;
            if (l1 != nullptr) {
                sum += l1->val;
                l1 = l1->next;
            }
            if (l2 != nullptr) {
                sum += l2->val;
                l2 = l2->next;
            }

            carry = sum /10;
            sum %= 10;
            curr->next = new ListNode(sum);
            curr = curr->next;
        }

        if (carry != 0) {
            curr->next = new ListNode(1);
        }
        curr = dummy->next;
        delete dummy;

        return curr;
    }
};

Time Complexity

$O(max(m,n))$

$m$ 為

$l1$ 的長度

$n$ 為

$l2$ 的長度

Space Complexity

$O(max(m,n))$

Miscellaneous

Related Coding Questions

(變形題) 67. Add Binary

class Solution {
public:
    string addBinary(string a, string b) {
        int i = a.size() - 1, j = b.size() - 1;
        int carry = 0;
        string answer;

        while (i >= 0 || j >= 0) {
            int sum = carry;
            if (i >= 0) {
                sum = sum + (a[i] - '0');
                --i;
            }
            if (j >= 0) {
                sum = sum + (b[j] - '0');
                --j;
            }

            carry = sum / 2;
            sum %= 2;
            answer.push_back(sum + '0');
        }

        if (carry != 0) {
            answer.push_back(carry + '0');
        }

        int left = 0, right = answer.size() - 1;
        while (left < right) {
            swap(answer[left++], answer[right--]);
        }

        return answer;
    }
};