Problem
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.
Examples
Example 1:
(2 -> 4 -> 3) + (5 -> 6 -> 4) => 7 -> 0 -> 8
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807
Solution
This problem is straight forward, but consider the cases:
- We should not forget the carry, especially when we have one more digit just for the carry. (e.g., 5 + 7 = 12, the ’1′ in 12 is due to the carry from last digit)
- Two numbers might not have same number of digits. (e.g., 1 + 123 are one digit and three digit numbers)
- As the link list are reversed that of number, we have taken care of adding it from right to left i.e. least significant digit to most significant digit.
Method 1 - OR in While Loop
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int carry = 0;
ListNode newHead = new ListNode(0);
ListNode l3 = newHead;
while (l1 != null || l2 != null) {
if (l1 != null) {
carry += l1.val;
l1 = l1.next;
}
if (l2 != null) {
carry += l2.val;
l2 = l2.next;
}
l3.next = new ListNode(carry % 10);
l3 = l3.next;
carry /= 10;
}
// important edge case
if (carry > 0) {
l3.next = new ListNode(carry);
}
return newHead.next;
}
Note that we are calling - if (p1!=null) and if(p2!=null) - we can optimize a little, but code is not that clean.
Method 2 - And in While Loop
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(-1);
ListNode dummyHead = dummy;
int extra = 0;
while (l1 != null && l2 != null) {
dummy.next = new ListNode((l1.val + l2.val + extra) % 10);
extra = l1.val + l2.val + extra > 9 ? 1 : 0;
dummy = dummy.next;
l1 = l1.next;
l2 = l2.next;
}
ListNode n = l1 != null ? l1 : l2;
while (n != null) {
dummy.next = new ListNode((n.val + extra) % 10);
extra = n.val + extra > 9 ? 1 : 0;
n = n.next;
dummy = dummy.next;
}
if (extra == 1) {
dummy.next = new ListNode(1);
}
return dummyHead.next;
}