题目
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.
示例
输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807
解题
解法一
和小时候学的竖式加法一样，从最小的一位，即从列表的第一个元素开始加起。遇到进位先保存，然后加入下一位即可。注意，两个列表不一定是相同长度的。
/**
* 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 tmp = null;
ListNode result = null;
int carry = 0;
while (l1 != null || l2 != null || carry != 0) {
int sum = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
carry = sum / 10;
ListNode node = new ListNode(sum % 10);
if (tmp == null) {
tmp = node;
result = tmp;
} else {
tmp.next = node;
tmp = tmp.next;
}
l1 = l1 == null ? null : l1.next;
l2 = l2 == null ? null : l2.next;
}
return result;
}
}
