Question

difficulty: mid
adj diff: ?

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

两种办法：

三次链表反转，具体代码参考 https://mar2ndx.github.io/2022/10/08/algo-2022-206-Reverse-Linked-List/
用Stack，更加推荐，代码简单。

方法2，如下：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    Stack<Integer> s1 = new Stack<Integer>();
    Stack<Integer> s2 = new Stack<Integer>();
    while (l1 != null) {
        s1.push(l1.val);
        l1 = l1.next;
    }
    while (l2 != null) {
        s2.push(l2.val);
        l2 = l2.next;
    }
    int carry = 0;
    ListNode ans = new ListNode();

    while (!s1.isEmpty() || !s2.isEmpty()) {
        int sum = carry + 
            (s1.isEmpty() ? 0 : s1.pop()) +
            (s2.isEmpty() ? 0 : s2.pop());
        ans.val = sum % 10;
        carry = sum / 10;
        ListNode tmp = new ListNode();
        tmp.next = ans;
        ans = tmp;
    }

    if (carry > 0) ans.val = carry;
    else ans = ans.next;
    return ans;
}