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.
Solution
- 声明变量进位(carry)和dummyHead方便保存链表的头部
- 遍历l1，l2两个链表直到l1和l2到最后并且进位已经为0
- 如果l1没有到最后，加上l1链表当前结点的值
- 同样，如果l1没有到最后，加上l1链表当前结点的值
- 加上进位值
- 更新进位值:
carry = val / 10 - 更新当前值:
val = val % 10
Code (Java)
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummyHead = new ListNode(-1);
ListNode cur = dummyHead;
int carry = 0;
while(l1 != null || l2 != null || carry != 0) {
int val = 0;
if(l1 != null) {
val += l1.val;
l1 = l1.next;
}
if(l2 != null) {
val += l2.val;
l2 = l2.next;
}
val += carry;
carry = val / 10;
val = val % 10;
cur.next = new ListNode(val);
cur = cur.next;
}
return dummyHead.next;
}
