【Leetcode】2. Add Two Numbers

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本文介绍了一种解决链表加法问题的方法，通过遍历两个逆序存储非负整数的链表并逐位相加，处理进位情况，最终返回结果链表。

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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 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.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

解法：这个题解法和add binary一样，定义一个变量carry来维持每一次进一的情况，定义一个变量value来累加每一次l1.val+l2.val+carry的值。注意考虑边界条件，1.当其中一个Linkedlist指针指到尽头(null)时，2.当两个Linkedlist的长度相等且指针都直到尽头(null)这时进一还有值(carry != 0)时。代码如下：

/**
 * 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 dummy = new ListNode(0);
        ListNode pre = dummy;
        int carry = 0;
        
        while (l1 != null || l2 != null){
            int value = 0;
            if (l1 != null){
                value += l1.val;
                l1 = l1.next;
            }
            if (l2 != null){
                value += l2.val;
                l2 = l2.next;
            }
            value += carry;
            carry = value / 10;
            value %= 10;
            pre.next = new ListNode(value);
            pre = pre.next;
        }
        if (carry != 0){
            pre.next = new ListNode(carry);
        }
        return dummy.next;
    }
}