Add Two Numbers II

Description

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:

Input: l1 = [7,2,4,3], l2 = [5,6,4] Output: [7,8,0,7]

Example 2:

Input: l1 = [2,4,3], l2 = [5,6,4] Output: [8,0,7]

Example 3:

Input: l1 = [0], l2 = [0] Output: [0]

Constraints:

• The number of nodes in each linked list is in the range [1, 100].
• 0 <= Node.val <= 9
• It is guaranteed that the list represents a number that does not have leading zeros.

Follow up: Could you solve it without reversing the input lists?

Code

Time Complexity: $O(N)$, Space Complexity: $O(1)$

結合 Reverse Linked List 和 Add Two Numbers 就可以解出。

/**
 * 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) {
        l1 = reverse_list(l1);
        l2 = reverse_list(l2);
 
        int carry = 0;
        ListNode* dummy = new ListNode(0);
        ListNode* curr = dummy;
        while(l1 || l2 || carry) {
            int sum = 0;
            if(l1) {
                sum += l1->val;
                l1 = l1->next;
            }
 
            if(l2) {
                sum += l2->val;
                l2 = l2->next;
            }
            sum += carry;
 
            curr->next = new ListNode(sum % 10);
            carry = sum / 10;
            curr = curr->next;
        }
        
        ListNode* res = reverse_list(dummy->next);
        return res;
    }
 
    ListNode* reverse_list(ListNode* head) {
        ListNode* prev = nullptr;
        ListNode* cur = head;
        while(cur) {
            ListNode* next = cur->next;
            cur->next = prev;
            prev = cur;
            cur = next;
        }
 
        return prev;
    }
};

Source

• Add Two Numbers II - LeetCode