2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

Solution1：遍历对应计算with 进位c

思路： MAINLY: sum += cur1.val + cur2.val + c;
当前结点=sum / 10; c = sum %10
Time Complexity: O(N) Space Complexity: O(1)

Solution2：Early stop version (faster)

Solution1 Code：

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode cur1 = l1;
        ListNode cur2 = l2;
        
        ListNode prev = dummy;
        int c = 0;
        while(cur1 != null || cur2 != null || c != 0) {
            int sum = c;
            if(cur1 != null) {
                sum += cur1.val;
                cur1 = cur1.next;
            }
            if(cur2 != null) {
                sum += cur2.val;
                cur2 = cur2.next;
            }
            ListNode node = new ListNode(sum % 10);
            c = sum / 10;
            
            prev.next = node;
            prev = node;
        }
        
        return dummy.next;
    }
}

Solution2 Code：

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        int c = 0;
        while(l1 != null && l2 != null) {
            int sum = c;
            sum += l1.val;
            sum += l2.val;
            ListNode node = new ListNode(sum % 10);
            c = sum / 10;
            
            cur.next = node;
            cur = cur.next;
            l1 = l1.next;
            l2 = l2.next;
        }
        
        while(c != 0) {
            int sum = c;
            if(l1 != null) sum += l1.val;
            if(l2 != null) sum += l2.val;
            ListNode node = new ListNode(sum % 10);
            c = sum / 10;
            cur.next = node;
            cur = cur.next;
            
            if(l1 != null) l1 = l1.next;
            if(l2 != null) l2 = l2.next;
        }
        if(l1 != null) cur.next = l1;
        else if(l2 != null) cur.next = l2;
        
        return dummy.next;
    }
}