PROBLEM DESCRIPTION
Given two numbers as LinkedList in reverseOrder (most significant digit as the right most digit), return a LinkedList (reversed) as the sum.
Example:
LinkedListOne: 2 -> 4 -> 7 -> 1
LinkedListTwo: 9 -> 4 -> 5Output: 1 -> 9 -> 2 -> 2
1742 + 946 = 2291
SOLUTION
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import java.util.*;
class Program {
public static class LinkedList {
public int value;
public LinkedList next;
public LinkedList(int value) {
this.value = value;
this.next = null;
}
}
public LinkedList insertAtEnd(LinkedList ll, int value){
if(ll == null){
return new LinkedList(value);
}
LinkedList temp = ll;
while(temp.next != null){
temp = temp.next;
}
LinkedList node = new LinkedList(value);
temp.next=node;
return ll;
}
public LinkedList sumOfLinkedLists(LinkedList linkedListOne, LinkedList linkedListTwo) {
LinkedList ans = null;
LinkedList x = linkedListOne;
LinkedList y = linkedListTwo;
int carry = 0;
while(x != null || y != null || carry != 0){ //since the linkedlist size can be different, we will need to keep adding until the longest one is complete. Also, after the last number, it's possible that the carry is > 0, in which case, we will need to add that digit also.
int first = 0;
int second = 0;
if(x != null){
first = x.value;
}
if(y != null){
second = y.value;
}
int total = first+second+carry;
int next = total%10; //the next digit will be sum of digits in the list at that place plus any carry from previous step
carry = total/10; //the carry for next step will be total/10
ans = insertAtEnd(ans, next);
if(x != null){ //if it's null, we have covered all digits of first list
x = x.next;
}
if(y != null){ //if it's null, we have covered all digits of second list
y = y.next;
}
}
return ans;
}
}
Another Way to Code
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class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummyHead = new ListNode(0);
ListNode tail = dummyHead;
int carry = 0;
while(l1 != null || l2 != null || carry != 0){
int x = (l1==null?0:l1.val);
int y = (l2==null?0:l2.val);
int total = x+y+carry;
tail.next = new ListNode(total%10);
tail = tail.next;
carry = total/10;
if(l1 != null) l1 = l1.next;
if(l2 != null) l2 = l2.next;
}
return dummyHead.next;
}
}