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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
eg:
2 4 3
+ 5 6 4
= 7 0 8
9 9 9
+ 8 1
= 7 1 0 1
维护dummy, curt Node, while循环来计算从高到低数位上的加法，每加完一次l1 = l1.next, l2 = l2.next. 要注意用一个carry来记录是否进位，如果carry == 0 则不需要，若carry > 0(只能为1）时，要让该位的运算加上carry. 还要记得处理最后一味存在进位的情况，比如
5
+ 5
= 0 1
像这样的话，l1, l2已经是null了，那么就要专门来说carry > 0的情况下还得在形成的LinkedList尾巴后面跟上一个val = 1的Node.
/**
* 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) {
if (l1 == null && l2 == null){
return null;
}
if (l1 == null){
return l2;
}
if (l2 == null){
return l1;
}
ListNode dummy = new ListNode(-1);
ListNode curt = dummy;
int carry = 0;
while (l1 != null && l2 != null){
int sum = l1.val + l2.val + carry;
int digit = sum % 10;
carry = sum / 10;
ListNode node = new ListNode(digit);
curt.next = node;
curt = curt.next;
l1 = l1.next;
l2 = l2.next;
}
while (l1 != null){
int sum = l1.val + carry;
int digit = sum % 10;
carry = sum / 10;
ListNode node = new ListNode(digit);
curt.next = node;
curt = curt.next;
l1 = l1.next;
}
while (l2 != null){
int sum = l2.val + carry;
int digit = sum % 10;
carry = sum / 10;ss
ListNode node = new ListNode(digit);
curt.next = node;
curt = curt.next;
l2 = l2.next;
}
if (carry > 0){
ListNode node = new ListNode(1);
curt.next = node;
}
return dummy.next;
}
}
