Question
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.
Let's Go!
Solve by using PREP.
- P - Two non-empty linked lists representing two non-negative integers.
- R - Return a linked list of the two numbers added
- E - Examples provided by question. (See Above)
- P - See Below
Before I knew about Node List...
var addTwoNumbers = function(l1, l2) {
const reverseOfL1 = reverseArr(l1)
const reverseOfL2 = reverseArr(l2)
let list = []
let carry = 0
for (let i=0; i<l1.length; i++) {
const sum = reverseOfL1[i] + reverseOfL2[i]
if (sum > 9) {
list.push(sum%10+carry)
carry = 1
} else {
list.push(sum+carry)
carry = 0
}
}
return list
}
function reverseArr(arr) {
return arr.reduce((newArr,num) => {
newArr.unshift(num)
return newArr
}, [])
}
Realizing my mistake and reading up on NodeList and Linked Lists. Answer below:
let addTwoNumbers = function(l1, l2) {
let dummyHead = new ListNode(0);
let p1 = l1;
let p2 = l2;
let current = dummyHead;
let carry = 0;
while (p1 !== null || p2 !== null) {
let x = (p1 !== null) ? p1.val : 0;
let y = (p2 !== null) ? p2.val : 0;
let sum = x + y + carry;
carry = Math.floor(sum / 10);
current.next = new ListNode(sum % 10);
current = current.next;
if (p1 !== null) {
p1 = p1.next;
}
if (p2 !== null) {
p2 = p2.next;
}
}
if (carry > 0) {
current.next = new ListNode(1);
}
return dummyHead.next;
};
Conclusion
& Remember... Happy coding, friends! =)
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