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.
题目大意
逆序的链表从低位进行相加，得出的结果也逆序输出，返回值是逆序结果链表的头结点。
解题思路
注意进位问题
Input: (9 -> 9 -> 9 -> 9 -> 9) + (1 -> )
Output: 0 -> 0 -> 0 -> 0 -> 0 -> 1
这里引入一个变量carry当做进制位。我们把两个链表每一个对应位置的节点进行相加，如果长度不同，那么短的链表后面就是0，如123的逆序就是3210以此类推。
当然不能忽略的就是如果循环结束，进制位不为0的话，还要新增加一个节点，把carry放进去。
代码：
/**
* 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) {
ListNode head = null, tail = null;
int carry = 0;
while (l1 != null || l2 != null) {
int n1 = l1 != null ? l1.val : 0; //当长度不等的时候短链表的节点值设为0
int n2 = l2 != null ? l2.val : 0;
/**
* 完整写法
int n1,n2;
if (l1 != null)
n1 = l1.val;
else
n1 = 0;
if (l2 != null)
n2 = l2.val;
else
n2 = 0;
**/
int sum = n1 + n2 + carry;
if (head == null) {
head = tail = new ListNode(sum % 10);
} else {
tail.next = new ListNode(sum % 10);
tail = tail.next;
}
carry = sum / 10; //进制位向下取整
if (l1 != null) {
l1 = l1.next;
}
if (l2 != null) {
l2 = l2.next;
}
}
if (carry > 0) { //如果循环结束，进制位不为0的话，还要新增加一个节点，把carry放进去
tail.next = new ListNode(carry);
}
return head;
}
}
Original: https://www.cnblogs.com/ancientlian/p/14294446.html
Author: Lian_tiam
Title: 2.Add Two Numbers——LeetCode
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