'LinkedList Add Two Numbers: LeetCode
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.
The linked list seems to be overwriting the nodes as far as I can tell. I did find the answer to this problem on GeeksforGeeks but I wanted help with figuring out what is wrong with my code. I also know my code is not the best in regards to optimization but any halp is accepted. Thanks!
/**
* 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 {
//ListNode head;
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int x = 0;
int y = 0;
int z = 1;
while(l1 != null){
x += z*l1.val;
z*=10;
l1 = l1.next;
}
z = 1;
while(l2 != null){
y += z*l2.val;
z*=10;
l2 = l2.next;
}
int sum = x + y;
ListNode node = new ListNode(0);
while(sum > 0){
int digit = sum % 10;
ListNode n = new ListNode(digit);
while(node.next != null){
node = node.next;
}
node.next = n;
sum = sum / 10;
}
return node;
}
}
Solution 1:[1]
I like your solution but you have bit incomplete logic.
class Solution {
//ListNode head;
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int x = 0;
int y = 0;
int z = 1;
while(l1 != null){
x += z*l1.val;
z*=10;
l1 = l1.next;
}
z = 1;
while(l2 != null){
y += z*l2.val;
z*=10;
l2 = l2.next;
}
int sum = x + y;
if (sum == 0) {
return new ListNode(0);
}
ListNode node = null, head = null;
while(sum > 0){
int digit = sum % 10;
ListNode n = new ListNode(digit);
if (node == null) {
head = node = n;
} else {
node.next = n;
node = node.next;
}
sum = sum / 10;
}
return head;
}
}
I just changed one or two things after int sum = x + y;
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | Alibek Taalaibek Uulu |