LeetCode-in-All
2. Add Two Numbers
Medium
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 contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example 1:
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
Example 2:
Input: l1 = [0], l2 = [0]
Output: [0]
Example 3:
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]
Constraints:
- The number of nodes in each linked list is in the range
[1, 100]. 0 <= Node.val <= 9- It is guaranteed that the list represents a number that does not have leading zeros.
Solution
using LeetCodeNet.Com_github_leetcode;
/**
* 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; }
* }
*/
public class Solution {
public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummyHead = new ListNode(0);
ListNode curr = dummyHead;
int carry = 0;
while (l1 != null || l2 != null) {
int x = (l1 != null) ? l1.val : 0;
int y = (l2 != null) ? l2.val : 0;
int sum = carry + x + y;
carry = sum / 10;
curr.next = new ListNode(sum % 10);
curr = curr.next;
if (l1 != null) l1 = l1.next;
if (l2 != null) l2 = l2.next;
}
if (carry > 0) {
curr.next = new ListNode(carry);
}
return dummyHead.next;
}
}