445 Add Two Numbers II

445. Add Two Numbers II

1. Question

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up: What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

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

Output: 7 -> 8 -> 0 ->7

2. Implementation

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode p1 = reverse(l1);
        ListNode p2 = reverse(l2);

        int sum = 0;
        ListNode dummy = new ListNode(0);
        ListNode curNode = dummy;

        while (p1 != null || p2 != null) {
            if (p1 != null) {
                sum += p1.val;
                p1 = p1.next;
            }

            if (p2 != null) {
                sum += p2.val;
                p2 = p2.next;
            }

            curNode.next = new ListNode(sum % 10);
            sum /= 10;
            curNode = curNode.next;
        }

        if (sum != 0) {
            curNode.next = new ListNode(sum);
        }

        return reverse(dummy.next);
    }

    public ListNode reverse (ListNode head) {
        ListNode curNode = head, preNode = null, nextNode = null;

        while (curNode != null) {
            nextNode = curNode.next;
            curNode.next = preNode;
            preNode = curNode;
            curNode = nextNode;
        }
        return preNode;
    }
}

3. Time & Space Complexity

时间复杂度O(m + n), m是第一个list的长度, n是第二个list的长度，空间复杂度O(1)