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 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.

给定两个表示两个非负整数的非空链表。这些数字以反向顺序存储，每个节点都包含一个数字。将这两个数相加，并以链表的形式返回它们的和。

假设这两个数字不包含任何前导零，除了数字0本身。

 

Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

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

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

 

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.

 

public class ListNode {
    int val;
    ListNode next;
    ListNode() {}
    ListNode(int val) { this.val = val; }
    ListNode(int val, ListNode next) { this.val = val; this.next = next; }
}

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode list = new ListNode(-1);
    ListNode current = list;
    int carry = 0;
    while (l1 != null || l2 != null) {
        int d1 = l1 == null ? 0 : l1.val;
        int d2 = l2 == null ? 0 : l2.val;
        int sum = d1 + d2 + carry;
        carry = sum >= 10 ? 1 : 0;
        current.next = new ListNode(sum % 10);
        current = current.next;
        if (l1 != null)
            l1 = l1.next;
        if (l2 != null)
            l2 = l2.next;
    }
    if (carry == 1)
        current.next = new ListNode(1);
    return list.next;
}

时间复杂度为O(n)，空间复杂度为O(n)。