Medium
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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.
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.
/**
* Definition for singly-linked list.
* 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; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
// 进位
int carry = 0;
// 结果，当前位
ListNode res, cur;
res = cur = new ListNode(0);
// 两个数的当前位
ListNode n1 = l1, n2 = l2;
while (n1 != null || n2 != null || carry != 0) {
// 当前位数相加，并加上进位
cur.val = (n1 != null ? n1.val : 0) + (n2 != null ? n2.val : 0) + carry;
// 清除进位
carry = 0;
// 如果当前产生了进位，则位数取个位，然后设置进位
if (cur.val >= 10) { cur.val -= 10; carry = 1; }
// 处理下一位
if (n1 != null) n1 = n1.next;
if (n2 != null) n2 = n2.next;
if (n1 != null || n2 != null || carry != 0) cur = cur.next = new ListNode(0);
}
return res;
}
}
